text stringlengths 198 433k | conversation_id int64 0 109k |
|---|---|
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given n integers. You need to choose a subset and put the chosen numbers in a beautiful rectangle (rectangular matrix). Each chosen number should occupy one of its rectangle cells, each cell must be filled with exactly one chosen number. Some of the n numbers may not be chosen.
A rectangle (rectangular matrix) is called beautiful if in each row and in each column all values are different.
What is the largest (by the total number of cells) beautiful rectangle you can construct? Print the rectangle itself.
Input
The first line contains n (1 β€ n β€ 4β
10^5). The second line contains n integers (1 β€ a_i β€ 10^9).
Output
In the first line print x (1 β€ x β€ n) β the total number of cells of the required maximum beautiful rectangle. In the second line print p and q (p β
q=x): its sizes. In the next p lines print the required rectangle itself. If there are several answers, print any.
Examples
Input
12
3 1 4 1 5 9 2 6 5 3 5 8
Output
12
3 4
1 2 3 5
3 1 5 4
5 6 8 9
Input
5
1 1 1 1 1
Output
1
1 1
1
Tags: brute force, combinatorics, constructive algorithms, data structures, greedy, math
Correct Solution:
```
from collections import Counter
from itertools import accumulate
from math import sqrt
from operator import itemgetter
import sys
n = int(input())
cnt = Counter(map(int, input().split()))
nums, counts = zip(*sorted(cnt.items(), key=itemgetter(1)))
acc = [0] + list(accumulate(counts))
area = 1
h, w = 1, 1
i = len(counts)
for y in range(int(sqrt(n)), 0, -1):
while i and counts[i-1] > y:
i -= 1
total = acc[i] + (len(counts) - i) * y
x = total // y
if y <= x and area < x * y:
h, w, area = y, x, x*y
ans = [[0]*w for _ in range(h)]
i = len(counts)-1
num, count = nums[i], min(h, counts[i])
for x in range(w):
for y in range(h):
ans[y][(x + y) % w] = num
count -= 1
if count == 0:
i -= 1
num, count = nums[i], h if h < counts[i] else counts[i]
print(area)
print(h, w)
for y in range(h):
sys.stdout.write(' '.join(map(str, ans[y])) + '\n')
```
| 94,500 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given n integers. You need to choose a subset and put the chosen numbers in a beautiful rectangle (rectangular matrix). Each chosen number should occupy one of its rectangle cells, each cell must be filled with exactly one chosen number. Some of the n numbers may not be chosen.
A rectangle (rectangular matrix) is called beautiful if in each row and in each column all values are different.
What is the largest (by the total number of cells) beautiful rectangle you can construct? Print the rectangle itself.
Input
The first line contains n (1 β€ n β€ 4β
10^5). The second line contains n integers (1 β€ a_i β€ 10^9).
Output
In the first line print x (1 β€ x β€ n) β the total number of cells of the required maximum beautiful rectangle. In the second line print p and q (p β
q=x): its sizes. In the next p lines print the required rectangle itself. If there are several answers, print any.
Examples
Input
12
3 1 4 1 5 9 2 6 5 3 5 8
Output
12
3 4
1 2 3 5
3 1 5 4
5 6 8 9
Input
5
1 1 1 1 1
Output
1
1 1
1
Tags: brute force, combinatorics, constructive algorithms, data structures, greedy, math
Correct Solution:
```
import sys
input = sys.stdin.readline
n=int(input())
A=list(map(int,input().split()))
from collections import Counter
C=Counter(A)
import math
MAXV=max(max(C.values()),int(math.sqrt(n)))
VCOUNT=[0]*(MAXV+1)
for v in C.values():
VCOUNT[v]+=1
SUM=n
from itertools import accumulate
ACC=list(accumulate(VCOUNT[::-1]))[::-1]
ANS=0
for i in range(MAXV,0,-1):
if SUM//i>=i:
if ANS<i*(SUM//i):
ANS=i*(SUM//i)
ANSX=i,(SUM//i)
SUM-=ACC[i]
print(ANS)
X,Y=ANSX[0],ANSX[1]
print(X,Y)
A=[[0]*Y for i in range(X)]
i=0
j=0
nowj=0
colored=0
same=0
LIST=list(C.most_common())
ind=0
while colored<ANS:
A[i][j],MAX=LIST[ind]
colored+=1
i+=1
j=(j+1)%Y
if i==X:
i=0
nowj+=1
j=nowj
same+=1
if same==min(X,MAX):
ind+=1
same=0
for a in A:
sys.stdout.write(" ".join(map(str,a))+"\n")
```
| 94,501 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given n integers. You need to choose a subset and put the chosen numbers in a beautiful rectangle (rectangular matrix). Each chosen number should occupy one of its rectangle cells, each cell must be filled with exactly one chosen number. Some of the n numbers may not be chosen.
A rectangle (rectangular matrix) is called beautiful if in each row and in each column all values are different.
What is the largest (by the total number of cells) beautiful rectangle you can construct? Print the rectangle itself.
Input
The first line contains n (1 β€ n β€ 4β
10^5). The second line contains n integers (1 β€ a_i β€ 10^9).
Output
In the first line print x (1 β€ x β€ n) β the total number of cells of the required maximum beautiful rectangle. In the second line print p and q (p β
q=x): its sizes. In the next p lines print the required rectangle itself. If there are several answers, print any.
Examples
Input
12
3 1 4 1 5 9 2 6 5 3 5 8
Output
12
3 4
1 2 3 5
3 1 5 4
5 6 8 9
Input
5
1 1 1 1 1
Output
1
1 1
1
Submitted Solution:
```
from math import sqrt
from collections import Counter
import sys
input = sys.stdin.readline
'''
for CASES in range(int(input())):
n, m = map(int, input().split())
n = int(input())
A = list(map(int, input().split()))
S = input().strip()
'''
inf = 100000000000000000 # 1e17
mod = 998244353
n = int(input())
A = list(map(int, input().split()))
D = Counter(A)
num = 0
maxx = 0
a = 0
b = 0
for i in range(1, int(sqrt(n)) + 1):
num = 0
for d in D.values():
num += min(i, d)
if num // i >= i and num // i * i > maxx:
maxx = num // i * i
a = i
b = num // i
print(maxx)
print(a, b)
A = []
D = D.most_common()
for key, d in D:
num = min(a, d)
for number in range(num):
A.append(key)
ANS = [[0] * (b) for _ in range(a)]
AIM = A[:maxx]
pos = 0
SET = set()
last = 0
for aim in AIM:
if last != aim:
SET.clear()
for i in range(b):
if ANS[pos][i] != 0:
continue
if i in SET:
continue
ANS[pos][i] = aim
SET.add(i)
pos += 1
pos %= a
last = aim
break
for i in range(a):
print(*ANS[i])
# the end
```
No
| 94,502 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given n integers. You need to choose a subset and put the chosen numbers in a beautiful rectangle (rectangular matrix). Each chosen number should occupy one of its rectangle cells, each cell must be filled with exactly one chosen number. Some of the n numbers may not be chosen.
A rectangle (rectangular matrix) is called beautiful if in each row and in each column all values are different.
What is the largest (by the total number of cells) beautiful rectangle you can construct? Print the rectangle itself.
Input
The first line contains n (1 β€ n β€ 4β
10^5). The second line contains n integers (1 β€ a_i β€ 10^9).
Output
In the first line print x (1 β€ x β€ n) β the total number of cells of the required maximum beautiful rectangle. In the second line print p and q (p β
q=x): its sizes. In the next p lines print the required rectangle itself. If there are several answers, print any.
Examples
Input
12
3 1 4 1 5 9 2 6 5 3 5 8
Output
12
3 4
1 2 3 5
3 1 5 4
5 6 8 9
Input
5
1 1 1 1 1
Output
1
1 1
1
Submitted Solution:
```
from math import sqrt
from collections import Counter
import sys
input = sys.stdin.readline
'''
for CASES in range(int(input())):
n, m = map(int, input().split())
n = int(input())
A = list(map(int, input().split()))
S = input().strip()
'''
inf = 100000000000000000 # 1e17
mod = 998244353
n = int(input())
A = list(map(int, input().split()))
D = Counter(A)
num = 0
maxx = 0
a = 0
b = 0
for i in range(1, int(sqrt(n)) + 1):
num = 0
for d in D.values():
num += min(i, d)
if num > maxx:
maxx = num
a = i
b = num // i
print(maxx)
print(a,b)
A = []
D = D.most_common()
for key, d in D:
num = min(a, d)
for number in range(num):
A.append(key)
ANS = [[0] * (b) for _ in range(a)]
AIM = A[:maxx]
pos = 0
SET = set()
last = 0
for aim in AIM:
if last != aim:
SET.clear()
for i in range(b):
if ANS[pos][i] != 0:
continue
if i in SET:
continue
ANS[pos][i] = aim
SET.add(i)
pos += 1
pos %= a
last = aim
break
for i in range(a):
print(*ANS[i])
# the end
```
No
| 94,503 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given n integers. You need to choose a subset and put the chosen numbers in a beautiful rectangle (rectangular matrix). Each chosen number should occupy one of its rectangle cells, each cell must be filled with exactly one chosen number. Some of the n numbers may not be chosen.
A rectangle (rectangular matrix) is called beautiful if in each row and in each column all values are different.
What is the largest (by the total number of cells) beautiful rectangle you can construct? Print the rectangle itself.
Input
The first line contains n (1 β€ n β€ 4β
10^5). The second line contains n integers (1 β€ a_i β€ 10^9).
Output
In the first line print x (1 β€ x β€ n) β the total number of cells of the required maximum beautiful rectangle. In the second line print p and q (p β
q=x): its sizes. In the next p lines print the required rectangle itself. If there are several answers, print any.
Examples
Input
12
3 1 4 1 5 9 2 6 5 3 5 8
Output
12
3 4
1 2 3 5
3 1 5 4
5 6 8 9
Input
5
1 1 1 1 1
Output
1
1 1
1
Submitted Solution:
```
from math import sqrt
from collections import Counter
import sys
input = sys.stdin.readline
'''
for CASES in range(int(input())):
n, m = map(int, input().split())
n = int(input())
A = list(map(int, input().split()))
S = input().strip()
'''
inf = 100000000000000000 # 1e17
mod = 998244353
n = int(input())
A = list(map(int, input().split()))
D = Counter(A)
num = 0
maxx = 0
a = 0
b = 0
for i in range(1, int(sqrt(n)) + 1):
num = 0
for d in D.values():
num += min(i, d)
if num//i>=i and num > maxx:
maxx = num
a = i
b = num // i
print(maxx)
print(a,b)
A = []
D = D.most_common()
for key, d in D:
num = min(a, d)
for number in range(num):
A.append(key)
ANS = [[0] * (b) for _ in range(a)]
AIM = A[:maxx]
pos = 0
SET = set()
last = 0
for aim in AIM:
if last != aim:
SET.clear()
for i in range(b):
if ANS[pos][i] != 0:
continue
if i in SET:
continue
ANS[pos][i] = aim
SET.add(i)
pos += 1
pos %= a
last = aim
break
for i in range(a):
print(*ANS[i])
# the end
```
No
| 94,504 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n monsters standing in a row numbered from 1 to n. The i-th monster has h_i health points (hp). You have your attack power equal to a hp and your opponent has his attack power equal to b hp.
You and your opponent are fighting these monsters. Firstly, you and your opponent go to the first monster and fight it till his death, then you and your opponent go the second monster and fight it till his death, and so on. A monster is considered dead if its hp is less than or equal to 0.
The fight with a monster happens in turns.
1. You hit the monster by a hp. If it is dead after your hit, you gain one point and you both proceed to the next monster.
2. Your opponent hits the monster by b hp. If it is dead after his hit, nobody gains a point and you both proceed to the next monster.
You have some secret technique to force your opponent to skip his turn. You can use this technique at most k times in total (for example, if there are two monsters and k=4, then you can use the technique 2 times on the first monster and 1 time on the second monster, but not 2 times on the first monster and 3 times on the second monster).
Your task is to determine the maximum number of points you can gain if you use the secret technique optimally.
Input
The first line of the input contains four integers n, a, b and k (1 β€ n β€ 2 β
10^5, 1 β€ a, b, k β€ 10^9) β the number of monsters, your attack power, the opponent's attack power and the number of times you can use the secret technique.
The second line of the input contains n integers h_1, h_2, ..., h_n (1 β€ h_i β€ 10^9), where h_i is the health points of the i-th monster.
Output
Print one integer β the maximum number of points you can gain if you use the secret technique optimally.
Examples
Input
6 2 3 3
7 10 50 12 1 8
Output
5
Input
1 1 100 99
100
Output
1
Input
7 4 2 1
1 3 5 4 2 7 6
Output
6
Tags: greedy, sortings
Correct Solution:
```
from sys import stdin, stdout
from math import ceil
n, a, b, k = map(int, stdin.readline().strip().split())
hlist = list(map(int, stdin.readline().strip().split()))
score = 0
lose_rems = []
for h in hlist:
t = h%(a+b)
if t == 0 or t>a:
# lose
# get the remainders when it is your turn
if t==0:
lose_rems.append(b)
else:
lose_rems.append(t-a)
else:
# win
score += 1
# its easier to finish smaller rems first
lose_rems.sort()
for rem in lose_rems:
skips = ceil(rem/a)
if k - skips >=0:
score += 1
k -= skips
else:
break
print(score)
```
| 94,505 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n monsters standing in a row numbered from 1 to n. The i-th monster has h_i health points (hp). You have your attack power equal to a hp and your opponent has his attack power equal to b hp.
You and your opponent are fighting these monsters. Firstly, you and your opponent go to the first monster and fight it till his death, then you and your opponent go the second monster and fight it till his death, and so on. A monster is considered dead if its hp is less than or equal to 0.
The fight with a monster happens in turns.
1. You hit the monster by a hp. If it is dead after your hit, you gain one point and you both proceed to the next monster.
2. Your opponent hits the monster by b hp. If it is dead after his hit, nobody gains a point and you both proceed to the next monster.
You have some secret technique to force your opponent to skip his turn. You can use this technique at most k times in total (for example, if there are two monsters and k=4, then you can use the technique 2 times on the first monster and 1 time on the second monster, but not 2 times on the first monster and 3 times on the second monster).
Your task is to determine the maximum number of points you can gain if you use the secret technique optimally.
Input
The first line of the input contains four integers n, a, b and k (1 β€ n β€ 2 β
10^5, 1 β€ a, b, k β€ 10^9) β the number of monsters, your attack power, the opponent's attack power and the number of times you can use the secret technique.
The second line of the input contains n integers h_1, h_2, ..., h_n (1 β€ h_i β€ 10^9), where h_i is the health points of the i-th monster.
Output
Print one integer β the maximum number of points you can gain if you use the secret technique optimally.
Examples
Input
6 2 3 3
7 10 50 12 1 8
Output
5
Input
1 1 100 99
100
Output
1
Input
7 4 2 1
1 3 5 4 2 7 6
Output
6
Tags: greedy, sortings
Correct Solution:
```
import sys
import math
input=sys.stdin.readline
n,a,b,k=map(int,input().split())
c=list(map(int,input().split()))
for j in range(n):
p=c[j]%(a+b)
if p==0:
c[j]=(a+b)
else:
c[j]=p
c.sort()
s=0
for j in c:
if j<=a:
s+=1
else:
q=math.ceil((j-a)/a)
if k>=q:
s+=1
k+=-q
else:
break
print(s)
```
| 94,506 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n monsters standing in a row numbered from 1 to n. The i-th monster has h_i health points (hp). You have your attack power equal to a hp and your opponent has his attack power equal to b hp.
You and your opponent are fighting these monsters. Firstly, you and your opponent go to the first monster and fight it till his death, then you and your opponent go the second monster and fight it till his death, and so on. A monster is considered dead if its hp is less than or equal to 0.
The fight with a monster happens in turns.
1. You hit the monster by a hp. If it is dead after your hit, you gain one point and you both proceed to the next monster.
2. Your opponent hits the monster by b hp. If it is dead after his hit, nobody gains a point and you both proceed to the next monster.
You have some secret technique to force your opponent to skip his turn. You can use this technique at most k times in total (for example, if there are two monsters and k=4, then you can use the technique 2 times on the first monster and 1 time on the second monster, but not 2 times on the first monster and 3 times on the second monster).
Your task is to determine the maximum number of points you can gain if you use the secret technique optimally.
Input
The first line of the input contains four integers n, a, b and k (1 β€ n β€ 2 β
10^5, 1 β€ a, b, k β€ 10^9) β the number of monsters, your attack power, the opponent's attack power and the number of times you can use the secret technique.
The second line of the input contains n integers h_1, h_2, ..., h_n (1 β€ h_i β€ 10^9), where h_i is the health points of the i-th monster.
Output
Print one integer β the maximum number of points you can gain if you use the secret technique optimally.
Examples
Input
6 2 3 3
7 10 50 12 1 8
Output
5
Input
1 1 100 99
100
Output
1
Input
7 4 2 1
1 3 5 4 2 7 6
Output
6
Tags: greedy, sortings
Correct Solution:
```
import math
n, a, b, k = [int(k) for k in input().split()]
h = []
points = 0
for currentMon in input().split():
remainingHealth = int(currentMon) % (a+b)
if remainingHealth == 0:
h.append(math.ceil(b/a))
else:
if remainingHealth <= a:
points += 1
else:
h.append(math.ceil(remainingHealth/a)-1)
h = sorted(h)
i = 0
while k > 0 and i < len(h):
if h[i] <= k:
k -= h[i]
points += 1
i += 1
else:
break
print(points)
```
| 94,507 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n monsters standing in a row numbered from 1 to n. The i-th monster has h_i health points (hp). You have your attack power equal to a hp and your opponent has his attack power equal to b hp.
You and your opponent are fighting these monsters. Firstly, you and your opponent go to the first monster and fight it till his death, then you and your opponent go the second monster and fight it till his death, and so on. A monster is considered dead if its hp is less than or equal to 0.
The fight with a monster happens in turns.
1. You hit the monster by a hp. If it is dead after your hit, you gain one point and you both proceed to the next monster.
2. Your opponent hits the monster by b hp. If it is dead after his hit, nobody gains a point and you both proceed to the next monster.
You have some secret technique to force your opponent to skip his turn. You can use this technique at most k times in total (for example, if there are two monsters and k=4, then you can use the technique 2 times on the first monster and 1 time on the second monster, but not 2 times on the first monster and 3 times on the second monster).
Your task is to determine the maximum number of points you can gain if you use the secret technique optimally.
Input
The first line of the input contains four integers n, a, b and k (1 β€ n β€ 2 β
10^5, 1 β€ a, b, k β€ 10^9) β the number of monsters, your attack power, the opponent's attack power and the number of times you can use the secret technique.
The second line of the input contains n integers h_1, h_2, ..., h_n (1 β€ h_i β€ 10^9), where h_i is the health points of the i-th monster.
Output
Print one integer β the maximum number of points you can gain if you use the secret technique optimally.
Examples
Input
6 2 3 3
7 10 50 12 1 8
Output
5
Input
1 1 100 99
100
Output
1
Input
7 4 2 1
1 3 5 4 2 7 6
Output
6
Tags: greedy, sortings
Correct Solution:
```
import math
n,a,b,k=map(int,input().split())
h=list(map(int,input().split()))
l=[]
c=0
for i in range(n):
r=h[i]%(a+b)
if r<=a and r!=0:c+=1
elif a>b:l.append(1)
else:l.append(math.ceil(r/a)-1 if r!=0 else math.ceil(b/a))
for x in sorted(l):
if k<=0 or x>k:break
else:c+=1;k-=x
print(c)
```
| 94,508 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n monsters standing in a row numbered from 1 to n. The i-th monster has h_i health points (hp). You have your attack power equal to a hp and your opponent has his attack power equal to b hp.
You and your opponent are fighting these monsters. Firstly, you and your opponent go to the first monster and fight it till his death, then you and your opponent go the second monster and fight it till his death, and so on. A monster is considered dead if its hp is less than or equal to 0.
The fight with a monster happens in turns.
1. You hit the monster by a hp. If it is dead after your hit, you gain one point and you both proceed to the next monster.
2. Your opponent hits the monster by b hp. If it is dead after his hit, nobody gains a point and you both proceed to the next monster.
You have some secret technique to force your opponent to skip his turn. You can use this technique at most k times in total (for example, if there are two monsters and k=4, then you can use the technique 2 times on the first monster and 1 time on the second monster, but not 2 times on the first monster and 3 times on the second monster).
Your task is to determine the maximum number of points you can gain if you use the secret technique optimally.
Input
The first line of the input contains four integers n, a, b and k (1 β€ n β€ 2 β
10^5, 1 β€ a, b, k β€ 10^9) β the number of monsters, your attack power, the opponent's attack power and the number of times you can use the secret technique.
The second line of the input contains n integers h_1, h_2, ..., h_n (1 β€ h_i β€ 10^9), where h_i is the health points of the i-th monster.
Output
Print one integer β the maximum number of points you can gain if you use the secret technique optimally.
Examples
Input
6 2 3 3
7 10 50 12 1 8
Output
5
Input
1 1 100 99
100
Output
1
Input
7 4 2 1
1 3 5 4 2 7 6
Output
6
Tags: greedy, sortings
Correct Solution:
```
import math
a = [int(x) for x in input().split()]
b = [int(x) for x in input().split()]
score = 0
turn = 0
for i in range(0,a[0]):
b[i]=b[i]%(a[1]+a[2])
if(b[i]==0): b[i] = (a[1]+a[2])
b.sort()
for i in range(0,a[0]):
if(b[i]<=a[1]):
score+=1
else:
b[i]-=a[1]
k = math.ceil(b[i]/a[1])
if(a[3]>=k):
a[3]-=k
score+=1
print(score)
```
| 94,509 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n monsters standing in a row numbered from 1 to n. The i-th monster has h_i health points (hp). You have your attack power equal to a hp and your opponent has his attack power equal to b hp.
You and your opponent are fighting these monsters. Firstly, you and your opponent go to the first monster and fight it till his death, then you and your opponent go the second monster and fight it till his death, and so on. A monster is considered dead if its hp is less than or equal to 0.
The fight with a monster happens in turns.
1. You hit the monster by a hp. If it is dead after your hit, you gain one point and you both proceed to the next monster.
2. Your opponent hits the monster by b hp. If it is dead after his hit, nobody gains a point and you both proceed to the next monster.
You have some secret technique to force your opponent to skip his turn. You can use this technique at most k times in total (for example, if there are two monsters and k=4, then you can use the technique 2 times on the first monster and 1 time on the second monster, but not 2 times on the first monster and 3 times on the second monster).
Your task is to determine the maximum number of points you can gain if you use the secret technique optimally.
Input
The first line of the input contains four integers n, a, b and k (1 β€ n β€ 2 β
10^5, 1 β€ a, b, k β€ 10^9) β the number of monsters, your attack power, the opponent's attack power and the number of times you can use the secret technique.
The second line of the input contains n integers h_1, h_2, ..., h_n (1 β€ h_i β€ 10^9), where h_i is the health points of the i-th monster.
Output
Print one integer β the maximum number of points you can gain if you use the secret technique optimally.
Examples
Input
6 2 3 3
7 10 50 12 1 8
Output
5
Input
1 1 100 99
100
Output
1
Input
7 4 2 1
1 3 5 4 2 7 6
Output
6
Tags: greedy, sortings
Correct Solution:
```
import math
n,a,b,k=map(int,input().split())
p=a+b
h=list(map(int,input().split()))
s=[0]*n
for i in range(n):
if h[i]%p==0:
h[i]=b
elif h[i]%p<=a:
s[i]=0
continue
else:
h[i]=h[i]%p-a
s[i]=math.ceil(h[i]/a)
s.sort()
ans=0
for i in range(n):
if s[i]==0:
ans+=1
elif k-s[i]>=0:
ans+=1
k-=s[i]
print(ans)
```
| 94,510 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n monsters standing in a row numbered from 1 to n. The i-th monster has h_i health points (hp). You have your attack power equal to a hp and your opponent has his attack power equal to b hp.
You and your opponent are fighting these monsters. Firstly, you and your opponent go to the first monster and fight it till his death, then you and your opponent go the second monster and fight it till his death, and so on. A monster is considered dead if its hp is less than or equal to 0.
The fight with a monster happens in turns.
1. You hit the monster by a hp. If it is dead after your hit, you gain one point and you both proceed to the next monster.
2. Your opponent hits the monster by b hp. If it is dead after his hit, nobody gains a point and you both proceed to the next monster.
You have some secret technique to force your opponent to skip his turn. You can use this technique at most k times in total (for example, if there are two monsters and k=4, then you can use the technique 2 times on the first monster and 1 time on the second monster, but not 2 times on the first monster and 3 times on the second monster).
Your task is to determine the maximum number of points you can gain if you use the secret technique optimally.
Input
The first line of the input contains four integers n, a, b and k (1 β€ n β€ 2 β
10^5, 1 β€ a, b, k β€ 10^9) β the number of monsters, your attack power, the opponent's attack power and the number of times you can use the secret technique.
The second line of the input contains n integers h_1, h_2, ..., h_n (1 β€ h_i β€ 10^9), where h_i is the health points of the i-th monster.
Output
Print one integer β the maximum number of points you can gain if you use the secret technique optimally.
Examples
Input
6 2 3 3
7 10 50 12 1 8
Output
5
Input
1 1 100 99
100
Output
1
Input
7 4 2 1
1 3 5 4 2 7 6
Output
6
Tags: greedy, sortings
Correct Solution:
```
n, a, b, k = list(map(int, input().split()))
ar = list(map(int, input().split()))
kek = []
ans = 0
for elem in ar:
x = elem % (a + b)
if x == 0 or x > a:
if x == 0:
x = a + b
kek.append((x + a - 1) // a - 1)
else:
ans += 1
kek.sort()
i = 0
while i < len(kek) and k > 0:
if k - kek[i] >= 0:
k -= kek[i]
i += 1
ans += 1
else:
break
print(ans)
```
| 94,511 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n monsters standing in a row numbered from 1 to n. The i-th monster has h_i health points (hp). You have your attack power equal to a hp and your opponent has his attack power equal to b hp.
You and your opponent are fighting these monsters. Firstly, you and your opponent go to the first monster and fight it till his death, then you and your opponent go the second monster and fight it till his death, and so on. A monster is considered dead if its hp is less than or equal to 0.
The fight with a monster happens in turns.
1. You hit the monster by a hp. If it is dead after your hit, you gain one point and you both proceed to the next monster.
2. Your opponent hits the monster by b hp. If it is dead after his hit, nobody gains a point and you both proceed to the next monster.
You have some secret technique to force your opponent to skip his turn. You can use this technique at most k times in total (for example, if there are two monsters and k=4, then you can use the technique 2 times on the first monster and 1 time on the second monster, but not 2 times on the first monster and 3 times on the second monster).
Your task is to determine the maximum number of points you can gain if you use the secret technique optimally.
Input
The first line of the input contains four integers n, a, b and k (1 β€ n β€ 2 β
10^5, 1 β€ a, b, k β€ 10^9) β the number of monsters, your attack power, the opponent's attack power and the number of times you can use the secret technique.
The second line of the input contains n integers h_1, h_2, ..., h_n (1 β€ h_i β€ 10^9), where h_i is the health points of the i-th monster.
Output
Print one integer β the maximum number of points you can gain if you use the secret technique optimally.
Examples
Input
6 2 3 3
7 10 50 12 1 8
Output
5
Input
1 1 100 99
100
Output
1
Input
7 4 2 1
1 3 5 4 2 7 6
Output
6
Tags: greedy, sortings
Correct Solution:
```
n,a,b,k=map(int,input().split())
l=list(map(int,input().split()))
for i in range(n) :
l[i]=l[i]%(a+b)
# print(l)
an=[]
c=0
for i in l :
if i>0 and i<=a :
c+=1
else :
if i==0 :
an.append(b)
else :
an.append(i-a)
an.sort()
# print(c)
# print(an)
le=len(an)
i=0
while (k>0 and i<le) :
if an[i]%a==0 :
if an[i]//a <=k :
k=k-an[i]//a
c+=1
else :
k=0
else :
if (an[i]//a)+1 <=k :
k=k-(an[i]//a)-1
c+=1
else :
k=0
i+=1
print(c)
```
| 94,512 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n monsters standing in a row numbered from 1 to n. The i-th monster has h_i health points (hp). You have your attack power equal to a hp and your opponent has his attack power equal to b hp.
You and your opponent are fighting these monsters. Firstly, you and your opponent go to the first monster and fight it till his death, then you and your opponent go the second monster and fight it till his death, and so on. A monster is considered dead if its hp is less than or equal to 0.
The fight with a monster happens in turns.
1. You hit the monster by a hp. If it is dead after your hit, you gain one point and you both proceed to the next monster.
2. Your opponent hits the monster by b hp. If it is dead after his hit, nobody gains a point and you both proceed to the next monster.
You have some secret technique to force your opponent to skip his turn. You can use this technique at most k times in total (for example, if there are two monsters and k=4, then you can use the technique 2 times on the first monster and 1 time on the second monster, but not 2 times on the first monster and 3 times on the second monster).
Your task is to determine the maximum number of points you can gain if you use the secret technique optimally.
Input
The first line of the input contains four integers n, a, b and k (1 β€ n β€ 2 β
10^5, 1 β€ a, b, k β€ 10^9) β the number of monsters, your attack power, the opponent's attack power and the number of times you can use the secret technique.
The second line of the input contains n integers h_1, h_2, ..., h_n (1 β€ h_i β€ 10^9), where h_i is the health points of the i-th monster.
Output
Print one integer β the maximum number of points you can gain if you use the secret technique optimally.
Examples
Input
6 2 3 3
7 10 50 12 1 8
Output
5
Input
1 1 100 99
100
Output
1
Input
7 4 2 1
1 3 5 4 2 7 6
Output
6
Submitted Solution:
```
n,a,b,k=map(int,input().split())
L=list(map(int,input().split()))
count=0
S=0
arr=[]
import math
for i in L:
if i<=a:
count+=1
elif i%(a+b)<=a and i%(a+b)!=0:
count+=1
elif i%(a+b)==0:
arr.append(math.ceil(b/a))
else:
arr.append(math.ceil((((i)%(a+b))-a)/a))
arr.sort()
for i in range(len(arr)):
if arr[i]<=k:
count+=1
k-=arr[i]
else:
break
print(count)
```
Yes
| 94,513 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n monsters standing in a row numbered from 1 to n. The i-th monster has h_i health points (hp). You have your attack power equal to a hp and your opponent has his attack power equal to b hp.
You and your opponent are fighting these monsters. Firstly, you and your opponent go to the first monster and fight it till his death, then you and your opponent go the second monster and fight it till his death, and so on. A monster is considered dead if its hp is less than or equal to 0.
The fight with a monster happens in turns.
1. You hit the monster by a hp. If it is dead after your hit, you gain one point and you both proceed to the next monster.
2. Your opponent hits the monster by b hp. If it is dead after his hit, nobody gains a point and you both proceed to the next monster.
You have some secret technique to force your opponent to skip his turn. You can use this technique at most k times in total (for example, if there are two monsters and k=4, then you can use the technique 2 times on the first monster and 1 time on the second monster, but not 2 times on the first monster and 3 times on the second monster).
Your task is to determine the maximum number of points you can gain if you use the secret technique optimally.
Input
The first line of the input contains four integers n, a, b and k (1 β€ n β€ 2 β
10^5, 1 β€ a, b, k β€ 10^9) β the number of monsters, your attack power, the opponent's attack power and the number of times you can use the secret technique.
The second line of the input contains n integers h_1, h_2, ..., h_n (1 β€ h_i β€ 10^9), where h_i is the health points of the i-th monster.
Output
Print one integer β the maximum number of points you can gain if you use the secret technique optimally.
Examples
Input
6 2 3 3
7 10 50 12 1 8
Output
5
Input
1 1 100 99
100
Output
1
Input
7 4 2 1
1 3 5 4 2 7 6
Output
6
Submitted Solution:
```
n,a,b,k = map(int,input().split())
L = list(map(int,input().split()))
for i in range(len(L)):
x = L[i] % (a + b)
if x != 0 and x >= 1 and x <= a:
L[i] = 0
else:
if x == 0:
x = a + b
x = x - a
if x % a == 0:
L[i] = x // a
else:
L[i] = x // a + 1
L.sort()
#print(*L)
ans = 0
for i in L:
if k >= i:
k -= i
ans += 1
print(ans)
```
Yes
| 94,514 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n monsters standing in a row numbered from 1 to n. The i-th monster has h_i health points (hp). You have your attack power equal to a hp and your opponent has his attack power equal to b hp.
You and your opponent are fighting these monsters. Firstly, you and your opponent go to the first monster and fight it till his death, then you and your opponent go the second monster and fight it till his death, and so on. A monster is considered dead if its hp is less than or equal to 0.
The fight with a monster happens in turns.
1. You hit the monster by a hp. If it is dead after your hit, you gain one point and you both proceed to the next monster.
2. Your opponent hits the monster by b hp. If it is dead after his hit, nobody gains a point and you both proceed to the next monster.
You have some secret technique to force your opponent to skip his turn. You can use this technique at most k times in total (for example, if there are two monsters and k=4, then you can use the technique 2 times on the first monster and 1 time on the second monster, but not 2 times on the first monster and 3 times on the second monster).
Your task is to determine the maximum number of points you can gain if you use the secret technique optimally.
Input
The first line of the input contains four integers n, a, b and k (1 β€ n β€ 2 β
10^5, 1 β€ a, b, k β€ 10^9) β the number of monsters, your attack power, the opponent's attack power and the number of times you can use the secret technique.
The second line of the input contains n integers h_1, h_2, ..., h_n (1 β€ h_i β€ 10^9), where h_i is the health points of the i-th monster.
Output
Print one integer β the maximum number of points you can gain if you use the secret technique optimally.
Examples
Input
6 2 3 3
7 10 50 12 1 8
Output
5
Input
1 1 100 99
100
Output
1
Input
7 4 2 1
1 3 5 4 2 7 6
Output
6
Submitted Solution:
```
from sys import stdin
from collections import deque
mod = 10**9 + 7
import sys
sys.setrecursionlimit(10**5)
from queue import PriorityQueue
# def rl():
# return [int(w) for w in stdin.readline().split()]
from bisect import bisect_right
from bisect import bisect_left
from collections import defaultdict
from math import sqrt,factorial,gcd,log2,inf,ceil
# map(int,input().split())
# # l = list(map(int,input().split()))
# from itertools import permutations
import heapq
# input = lambda: sys.stdin.readline().rstrip()
input = lambda : sys.stdin.readline().rstrip()
from sys import stdin, stdout
from heapq import heapify, heappush, heappop
# t = int(input())
#
# for _ in range(t):
#
# n = int(input())
# s = input()
#
# seti = set()
# x,y = 0,0
# hash = defaultdict(int)
# seti.add((x,y))
# hash[(x,y)] = 0
# mini = inf
# ans1 = -1
# ans2 = -1
# for i in range(n):
# if s[i] == 'L':
# x-=1
# if s[i] == 'R':
# x+=1
# if s[i] == 'U':
# y+=1
# if s[i] == 'D':
# y-=1
# # print((x,y))
# if (x,y) in seti:
#
# z = hash[(x,y)]
# if z == 0:
# len = i-z+1
# else:
# len = i-z
# mini = min(mini,len)
# # print(len,z)
# if mini == len:
# if z == 0:
# ans1 = z+1
# ans2 = i+1
# else:
# ans1 = z+2
# ans2 = i+1
# hash[(x,y)] = i
# hash[(x,y)] = i
# seti.add((x,y))
#
# if ans1 == -1 or ans2 == -1:
# print(-1)
# else:
# print(ans1,ans2)
#
#
n,a,b,k = map(int,input().split())
l = list(map(int,input().split()))
for i in range(n):
z = l[i]%(a+b)
if z == 0:
z = inf
l[i] = z
l.sort()
ans = 0
# print(l)
for i in range(n):
if a>=l[i]:
ans+=1
else:
if l[i]!=inf:
z = ceil(l[i]/a) - 1
if k>=z:
k-=z
ans+=1
else:
z = ceil((a+b)/a) - 1
if k>=z:
k-=z
ans+=1
print(ans)
```
Yes
| 94,515 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n monsters standing in a row numbered from 1 to n. The i-th monster has h_i health points (hp). You have your attack power equal to a hp and your opponent has his attack power equal to b hp.
You and your opponent are fighting these monsters. Firstly, you and your opponent go to the first monster and fight it till his death, then you and your opponent go the second monster and fight it till his death, and so on. A monster is considered dead if its hp is less than or equal to 0.
The fight with a monster happens in turns.
1. You hit the monster by a hp. If it is dead after your hit, you gain one point and you both proceed to the next monster.
2. Your opponent hits the monster by b hp. If it is dead after his hit, nobody gains a point and you both proceed to the next monster.
You have some secret technique to force your opponent to skip his turn. You can use this technique at most k times in total (for example, if there are two monsters and k=4, then you can use the technique 2 times on the first monster and 1 time on the second monster, but not 2 times on the first monster and 3 times on the second monster).
Your task is to determine the maximum number of points you can gain if you use the secret technique optimally.
Input
The first line of the input contains four integers n, a, b and k (1 β€ n β€ 2 β
10^5, 1 β€ a, b, k β€ 10^9) β the number of monsters, your attack power, the opponent's attack power and the number of times you can use the secret technique.
The second line of the input contains n integers h_1, h_2, ..., h_n (1 β€ h_i β€ 10^9), where h_i is the health points of the i-th monster.
Output
Print one integer β the maximum number of points you can gain if you use the secret technique optimally.
Examples
Input
6 2 3 3
7 10 50 12 1 8
Output
5
Input
1 1 100 99
100
Output
1
Input
7 4 2 1
1 3 5 4 2 7 6
Output
6
Submitted Solution:
```
class Monster:
def __init__(self, vidamonstruo):
self.hp= vidamonstruo
class Fighter:
def __init__(self, ataque, numerodetecnicas):
self.attack_power= ataque
self.number_techniques= numerodetecnicas
def fight(self,other,monster):
ataquepersonaje=self.attack_power
ataqueenemigo=other.attack_power
vidamonstruo= monster.hp
ataquetotal=(ataquepersonaje+ataqueenemigo)
numerotecnicasfinal= (vidamonstruo%ataquetotal)
if(numerotecnicasfinal ==0):
return(((ataquetotal+ataquepersonaje-1)//ataquepersonaje)-1)
else:
return(((numerotecnicasfinal+ataquepersonaje-1)//ataquepersonaje)-1)
def funcionprincipal(n,a,b,k,listademonstruos):
puntostotales = 0
jugador1, jugador2 = Fighter(a,k), Fighter(b,0)
listadetecnicas= []
for x in range(n):
monstruo=Monster(listademonstruos[x])
listadetecnicas.append(jugador1.fight(jugador2, monstruo))
listadetecnicas.sort()
for tecnica in listadetecnicas:
if((k-tecnica) < 0):
return puntostotales
else:
k-=tecnica
puntostotales+=1
return puntostotales
elementos=list(map(int,input().split()))
monstruos= list(map(int,input().split()))
print(funcionprincipal(elementos[0],elementos[1],elementos[2],elementos[3],monstruos))
```
Yes
| 94,516 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n monsters standing in a row numbered from 1 to n. The i-th monster has h_i health points (hp). You have your attack power equal to a hp and your opponent has his attack power equal to b hp.
You and your opponent are fighting these monsters. Firstly, you and your opponent go to the first monster and fight it till his death, then you and your opponent go the second monster and fight it till his death, and so on. A monster is considered dead if its hp is less than or equal to 0.
The fight with a monster happens in turns.
1. You hit the monster by a hp. If it is dead after your hit, you gain one point and you both proceed to the next monster.
2. Your opponent hits the monster by b hp. If it is dead after his hit, nobody gains a point and you both proceed to the next monster.
You have some secret technique to force your opponent to skip his turn. You can use this technique at most k times in total (for example, if there are two monsters and k=4, then you can use the technique 2 times on the first monster and 1 time on the second monster, but not 2 times on the first monster and 3 times on the second monster).
Your task is to determine the maximum number of points you can gain if you use the secret technique optimally.
Input
The first line of the input contains four integers n, a, b and k (1 β€ n β€ 2 β
10^5, 1 β€ a, b, k β€ 10^9) β the number of monsters, your attack power, the opponent's attack power and the number of times you can use the secret technique.
The second line of the input contains n integers h_1, h_2, ..., h_n (1 β€ h_i β€ 10^9), where h_i is the health points of the i-th monster.
Output
Print one integer β the maximum number of points you can gain if you use the secret technique optimally.
Examples
Input
6 2 3 3
7 10 50 12 1 8
Output
5
Input
1 1 100 99
100
Output
1
Input
7 4 2 1
1 3 5 4 2 7 6
Output
6
Submitted Solution:
```
import sys
import os
from io import IOBase, BytesIO
# import heapq
import math
# import collections
# import itertools
# import bisect
mod = 10 ** 9 + 7
pie = 3.1415926536
# import resource
# resource.setrlimit(resource.RLIMIT_STACK, [0x100000000, resource.RLIM_INFINITY])
# import threading
# threading.stack_size(2**27)
# import sys
# sys.setrecursionlimit(10**6)
# fact=[1]
# for i in range(1,1000001):
# fact.append((fact[-1]*i)%mod)
# ifact=[0]*1000001
# ifact[1000000]=pow(fact[1000000],mod-2,mod)
# for i in range(1000000,0,-1):
# ifact[i-1]=(i*ifact[i])%mod
# from random import randint as rn
# from Queue import Queue as Q
def modinv(n, p):
return pow(n, p-2, p)
def ncr(n, r, p): # for using this uncomment the lines calculating fact and ifact
t = ((fact[n])*((ifact[r]*ifact[n-r]) % p)) % p
return t
def ain(): # takes array as input
return list(map(int, sin().split()))
def sin():
return input().strip()
def GCD(x, y):
while(y):
x, y = y, x % y
return x
def read2DIntArray(row, col):
arr = []
for i in range(0, row):
temp = list(map(int, sin().split()))
arr.append(temp)
return arr
def read2DCharArray(row, col):
arr = []
for i in range(0, row):
temp = str(sin())
arr.append(temp)
return arr
"""****************** SMALLEST NO. BY REARRANGING DIGITS OF n (WITHOUT TRAILING ZEROS) *********************"""
def smallestNumber(n):
lst = list(str(n))
lst.sort()
tmp = ""
for i, n in enumerate(lst):
if (n != '0'):
tmp = lst.pop(i)
break
return str(tmp) + ''.join(lst)
"""*********************** GENERATE ALL PRIME NUMBERS SMALLER THAN OR EQUAL TO n ***************************"""
def SieveOfEratosthenes(n):
prime = [True for i in range(n + 1)]
p = 2
while (p * p <= n):
if (prime[p] == True):
for i in range(p * 2, n + 1, p):
prime[i] = False
p += 1
prime[0] = False
prime[1] = False
lst = []
for p in range(n + 1):
if prime[p]:
lst.append(p)
return lst
"""*************************************** FIND nCr ********************************************************"""
def nCr(n, r):
a = 1
b = 1
c = 1
for i in range(1, n + 1):
c *= i
for i in range(1, r + 1):
b *= i
for i in range(1, n - r + 1):
a *= i
return (c // (a * b))
"""************************* GET PRIME FACTORS AND THEIR POWERS FOR AN INTEGER *****************************"""
def sieveOfEratosthenes1(N, s):
prime = [False] * (N+1)
for i in range(2, N+1, 2):
s[i] = 2
for i in range(3, N+1, 2):
if (prime[i] == False):
s[i] = i
for j in range(i, int(N / i) + 1, 2):
if (prime[i*j] == False):
prime[i*j] = True
s[i * j] = i
def generatePrimeFactors(N):
s = [0] * (N+1)
sieveOfEratosthenes1(N, s)
# print("Factor Power")
curr = s[N]
cnt = 1
factors = []
power = []
while (N > 1):
N //= s[N]
if (curr == s[N]):
cnt += 1
continue
# curr is factor and cnt in the power of this factor
factors.append(curr)
power.append(cnt)
curr = s[N]
cnt = 1
return factors, power
"""----------------------------------------------MAIN------------------------------------------------------"""
def main():
n, a, b, k = map(int, sin().split())
h = ain()
ans = 0
for hp in h:
if hp <= a:
ans += 1
else:
rem = hp % (a + b)
if rem == 0:
rem = a+b
while k > 0 and rem > 0:
k -= 1
rem -= a
if rem <= 0:
ans += 1
else:
if rem <= a:
ans += 1
else:
while k > 0 and rem > 0:
k -= 1
rem -= a
if rem <= 0:
ans += 1
print(ans+1)
"""--------------------------------------------------------------------------------------------------------"""
# Python 2 and 3 footer by Pajenegod and j1729
py2 = round(0.5)
if py2:
from future_builtins import ascii, filter, hex, map, oct, zip
range = xrange
BUFSIZE = 8192
class FastIO(BytesIO):
newlines = 0
def __init__(self, file):
self._file = file
self._fd = file.fileno()
self.writable = "x" in file.mode or "w" in file.mode
self.write = super(FastIO, self).write if self.writable else None
def _fill(self):
s = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.seek((self.tell(), self.seek(0, 2),
super(FastIO, self).write(s))[0])
return s
def read(self):
while self._fill():
pass
return super(FastIO, self).read()
def readline(self):
while self.newlines == 0:
s = self._fill()
self.newlines = s.count(b"\n") + (not s)
self.newlines -= 1
return super(FastIO, self).readline()
def flush(self):
if self.writable:
os.write(self._fd, self.getvalue())
self.truncate(0), self.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
if py2:
self.write = self.buffer.write
self.read = self.buffer.read
self.readline = self.buffer.readline
else:
self.write = lambda s: self.buffer.write(s.encode('ascii'))
self.read = lambda: self.buffer.read().decode('ascii')
self.readline = lambda: self.buffer.readline().decode('ascii')
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
def input(): return sys.stdin.readline().rstrip('\r\n')
if __name__ == '__main__':
main()
# threading.Thread(target=main).start()
```
No
| 94,517 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n monsters standing in a row numbered from 1 to n. The i-th monster has h_i health points (hp). You have your attack power equal to a hp and your opponent has his attack power equal to b hp.
You and your opponent are fighting these monsters. Firstly, you and your opponent go to the first monster and fight it till his death, then you and your opponent go the second monster and fight it till his death, and so on. A monster is considered dead if its hp is less than or equal to 0.
The fight with a monster happens in turns.
1. You hit the monster by a hp. If it is dead after your hit, you gain one point and you both proceed to the next monster.
2. Your opponent hits the monster by b hp. If it is dead after his hit, nobody gains a point and you both proceed to the next monster.
You have some secret technique to force your opponent to skip his turn. You can use this technique at most k times in total (for example, if there are two monsters and k=4, then you can use the technique 2 times on the first monster and 1 time on the second monster, but not 2 times on the first monster and 3 times on the second monster).
Your task is to determine the maximum number of points you can gain if you use the secret technique optimally.
Input
The first line of the input contains four integers n, a, b and k (1 β€ n β€ 2 β
10^5, 1 β€ a, b, k β€ 10^9) β the number of monsters, your attack power, the opponent's attack power and the number of times you can use the secret technique.
The second line of the input contains n integers h_1, h_2, ..., h_n (1 β€ h_i β€ 10^9), where h_i is the health points of the i-th monster.
Output
Print one integer β the maximum number of points you can gain if you use the secret technique optimally.
Examples
Input
6 2 3 3
7 10 50 12 1 8
Output
5
Input
1 1 100 99
100
Output
1
Input
7 4 2 1
1 3 5 4 2 7 6
Output
6
Submitted Solution:
```
n,a,b,k = map(int,input().split(" "))
arr = [[int(x),0] for x in input().split(" ")]
for i in range(n):
var = arr[i][0]
if var %(a+b) <= a:
if var%(a+b)!=0:
arr[i][1] = -1
else:
arr[i][1] = b//a
if b//a < b/a: arr[i][1] += 1
else:
reduce = var%(a+b)
times = 0
if reduce>=a and b>a:
arr[i][1] = reduce//a
if reduce//a < reduce/a: arr[i][1] += 1
arr[i][1] -= 1
elif reduce<=b:
arr[i][1] = reduce//a
if reduce//a < reduce/a: arr[i][1] += 1
else:
arr[i][1] = reduce//a
if reduce//a < reduce/a: arr[i][1] += 1
arr.sort(key = lambda x:x[1])
ans = 0
for var in arr:
if var[1]==-1: ans += 1
elif k-var[1]>=0:
k-=var[1]
ans += 1
print(ans)
```
No
| 94,518 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n monsters standing in a row numbered from 1 to n. The i-th monster has h_i health points (hp). You have your attack power equal to a hp and your opponent has his attack power equal to b hp.
You and your opponent are fighting these monsters. Firstly, you and your opponent go to the first monster and fight it till his death, then you and your opponent go the second monster and fight it till his death, and so on. A monster is considered dead if its hp is less than or equal to 0.
The fight with a monster happens in turns.
1. You hit the monster by a hp. If it is dead after your hit, you gain one point and you both proceed to the next monster.
2. Your opponent hits the monster by b hp. If it is dead after his hit, nobody gains a point and you both proceed to the next monster.
You have some secret technique to force your opponent to skip his turn. You can use this technique at most k times in total (for example, if there are two monsters and k=4, then you can use the technique 2 times on the first monster and 1 time on the second monster, but not 2 times on the first monster and 3 times on the second monster).
Your task is to determine the maximum number of points you can gain if you use the secret technique optimally.
Input
The first line of the input contains four integers n, a, b and k (1 β€ n β€ 2 β
10^5, 1 β€ a, b, k β€ 10^9) β the number of monsters, your attack power, the opponent's attack power and the number of times you can use the secret technique.
The second line of the input contains n integers h_1, h_2, ..., h_n (1 β€ h_i β€ 10^9), where h_i is the health points of the i-th monster.
Output
Print one integer β the maximum number of points you can gain if you use the secret technique optimally.
Examples
Input
6 2 3 3
7 10 50 12 1 8
Output
5
Input
1 1 100 99
100
Output
1
Input
7 4 2 1
1 3 5 4 2 7 6
Output
6
Submitted Solution:
```
#1. Resuelva el problema usando python 3 y POO.
#2. Cree la clase Fighter con al menos los atributos number_techniques y attack_power .
#3. Cree la clase Monster con al menos el atributo hp.
#4. Cree el mΓ©todo Fighter.fight(self, other, monster) que retorne el nΓΊmero de tΓ©cnicas que debe usar self para obtener un punto.
class Fighter:
def __init__(self, number_techniques,attack_power):
self.number_techniques = number_techniques
self.attack_power = attack_power
#Esta funcion nos da la cantidad de habilidad que usaremos para ganar un punto por matar, se basa en la vida minima del monstruo antes de morir
def fight(self, other, monster):
self = self.attack_power
other= other.attack_power
hp_monstruo = monster.hp
hp= hp_monstruo % (self+other)
#puntos de vida del monstruo previo a atacar sin habilidad
if (hp==0):
puntos_de_vida= self+other #suma de ataque de poder self y other
else:
puntos_de_vida=hp
#Ataque final sin habilidad
puntos_de_vida-= self
if (puntos_de_vida > 0): #no murio asi que ocupo habilidad
return((puntos_de_vida + self -1 ) // self) #Ataques que quedan
else: # murio sin habilidad
return 0
class Monster:
def __init__(self, hp):
self.hp = hp
#inputs
input1 = (input()).split(' ') # 4 enteros} n: cant monstruos, a: poder ataque, b: poder ataque oponente y k:cant veces que puedo ocupar tecnica secreta
input2 = (input()).split(' ') # Puntos de vida i-nesimo monstuo
#Lista datos obtenidos de metodo fight
lista_healt=list (input2)
lista_tec=[]
puntos=0
fight_usuario_a = Fighter(int(input1[1]), int(input1[3]))
fight_usuario_b = Fighter(int(input1[2]), 0)
n = int(input1[0]) #n: cant monstruo
for contador in range(n):
lista_tec.append(fight_usuario_a.fight(fight_usuario_b, Monster(int(lista_healt[contador]))))
lista_tec.sort()
#calculo puntaje
for contador2 in lista_tec:
if ( contador2 !=0):
if (input1[3]>=contador2):
puntos += 1
input1[3]-=contador2
else:
puntos+=1
print(puntos)
#ideas:-http://docs.python.org.ar/tutorial/3/classes.html
```
No
| 94,519 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n monsters standing in a row numbered from 1 to n. The i-th monster has h_i health points (hp). You have your attack power equal to a hp and your opponent has his attack power equal to b hp.
You and your opponent are fighting these monsters. Firstly, you and your opponent go to the first monster and fight it till his death, then you and your opponent go the second monster and fight it till his death, and so on. A monster is considered dead if its hp is less than or equal to 0.
The fight with a monster happens in turns.
1. You hit the monster by a hp. If it is dead after your hit, you gain one point and you both proceed to the next monster.
2. Your opponent hits the monster by b hp. If it is dead after his hit, nobody gains a point and you both proceed to the next monster.
You have some secret technique to force your opponent to skip his turn. You can use this technique at most k times in total (for example, if there are two monsters and k=4, then you can use the technique 2 times on the first monster and 1 time on the second monster, but not 2 times on the first monster and 3 times on the second monster).
Your task is to determine the maximum number of points you can gain if you use the secret technique optimally.
Input
The first line of the input contains four integers n, a, b and k (1 β€ n β€ 2 β
10^5, 1 β€ a, b, k β€ 10^9) β the number of monsters, your attack power, the opponent's attack power and the number of times you can use the secret technique.
The second line of the input contains n integers h_1, h_2, ..., h_n (1 β€ h_i β€ 10^9), where h_i is the health points of the i-th monster.
Output
Print one integer β the maximum number of points you can gain if you use the secret technique optimally.
Examples
Input
6 2 3 3
7 10 50 12 1 8
Output
5
Input
1 1 100 99
100
Output
1
Input
7 4 2 1
1 3 5 4 2 7 6
Output
6
Submitted Solution:
```
import sys,math
input = sys.stdin.readline
n,a,b,k = map(int,input().split())
arr = list(map(int,input().split()))
pts,ans = [],0
for i in arr:
if(i<=a):
pts.append(0)
elif(i<a+b):
pts.append(i-a)
else:
p = i%(a+b)
if(p==0):
pts.append(math.ceil(b/a))
elif(p<=a):
pts.append(0)
else:
pts.append(math.ceil((p-a)/a))
temp = 0
pts.sort()
for i in pts:
temp+=i
if(temp>k):
break
ans+=1
print(ans)
```
No
| 94,520 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A subway scheme, classic for all Berland cities is represented by a set of n stations connected by n passages, each of which connects exactly two stations and does not pass through any others. Besides, in the classic scheme one can get from any station to any other one along the passages. The passages can be used to move in both directions. Between each pair of stations there is no more than one passage.
Berland mathematicians have recently proved a theorem that states that any classic scheme has a ringroad. There can be only one ringroad. In other words, in any classic scheme one can find the only scheme consisting of stations (where any two neighbouring ones are linked by a passage) and this cycle doesn't contain any station more than once.
This invention had a powerful social impact as now the stations could be compared according to their distance from the ringroad. For example, a citizen could say "I live in three passages from the ringroad" and another one could reply "you loser, I live in one passage from the ringroad". The Internet soon got filled with applications that promised to count the distance from the station to the ringroad (send a text message to a short number...).
The Berland government decided to put an end to these disturbances and start to control the situation. You are requested to write a program that can determine the remoteness from the ringroad for each station by the city subway scheme.
Input
The first line contains an integer n (3 β€ n β€ 3000), n is the number of stations (and trains at the same time) in the subway scheme. Then n lines contain descriptions of the trains, one per line. Each line contains a pair of integers xi, yi (1 β€ xi, yi β€ n) and represents the presence of a passage from station xi to station yi. The stations are numbered from 1 to n in an arbitrary order. It is guaranteed that xi β yi and that no pair of stations contain more than one passage. The passages can be used to travel both ways. It is guaranteed that the given description represents a classic subway scheme.
Output
Print n numbers. Separate the numbers by spaces, the i-th one should be equal to the distance of the i-th station from the ringroad. For the ringroad stations print number 0.
Examples
Input
4
1 3
4 3
4 2
1 2
Output
0 0 0 0
Input
6
1 2
3 4
6 4
2 3
1 3
3 5
Output
0 0 0 1 1 2
Tags: dfs and similar, graphs
Correct Solution:
```
import sys
import math
n=int(input())
e=[[] for _ in range(n)]
for _ in range(n):
x,y=map(int,input().split())
e[x-1].append(y-1)
e[y-1].append(x-1)
d=[len(x) for x in e]
q=[i for i in range(n) if d[i]==1]
c=[0]*n
for u in q:
c[u]=n
for v in e[u]:
d[v]-=1
if d[v]==1:
q.append(v)
q.extend(i for i in range(n) if not c[i])
for u in q[::-1]:
for v in e[u]:
c[v]=min(c[v],c[u]+1)
print(" ".join(map(str,c)))
```
| 94,521 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A subway scheme, classic for all Berland cities is represented by a set of n stations connected by n passages, each of which connects exactly two stations and does not pass through any others. Besides, in the classic scheme one can get from any station to any other one along the passages. The passages can be used to move in both directions. Between each pair of stations there is no more than one passage.
Berland mathematicians have recently proved a theorem that states that any classic scheme has a ringroad. There can be only one ringroad. In other words, in any classic scheme one can find the only scheme consisting of stations (where any two neighbouring ones are linked by a passage) and this cycle doesn't contain any station more than once.
This invention had a powerful social impact as now the stations could be compared according to their distance from the ringroad. For example, a citizen could say "I live in three passages from the ringroad" and another one could reply "you loser, I live in one passage from the ringroad". The Internet soon got filled with applications that promised to count the distance from the station to the ringroad (send a text message to a short number...).
The Berland government decided to put an end to these disturbances and start to control the situation. You are requested to write a program that can determine the remoteness from the ringroad for each station by the city subway scheme.
Input
The first line contains an integer n (3 β€ n β€ 3000), n is the number of stations (and trains at the same time) in the subway scheme. Then n lines contain descriptions of the trains, one per line. Each line contains a pair of integers xi, yi (1 β€ xi, yi β€ n) and represents the presence of a passage from station xi to station yi. The stations are numbered from 1 to n in an arbitrary order. It is guaranteed that xi β yi and that no pair of stations contain more than one passage. The passages can be used to travel both ways. It is guaranteed that the given description represents a classic subway scheme.
Output
Print n numbers. Separate the numbers by spaces, the i-th one should be equal to the distance of the i-th station from the ringroad. For the ringroad stations print number 0.
Examples
Input
4
1 3
4 3
4 2
1 2
Output
0 0 0 0
Input
6
1 2
3 4
6 4
2 3
1 3
3 5
Output
0 0 0 1 1 2
Tags: dfs and similar, graphs
Correct Solution:
```
from collections import *
read_line = lambda: [int(i) for i in input().split(' ')]
n = read_line()[0]
g, deg = [[] for i in range(n)], [0] * n
for i in range(n):
a, b = [v - 1 for v in read_line()]
g[a].append(b)
g[b].append(a)
deg[a] += 1
deg[b] += 1
ans = [0] * n
q = deque(v for v in range(n) if deg[v] == 1)
while q:
v = q.popleft()
ans[v] = None
for u in g[v]:
deg[u] -= 1
if deg[u] == 1:
q.append(u)
q = deque(v for v in range(n) if ans[v] == 0)
while q:
v = q.popleft()
for u in g[v]:
if ans[u] is None:
ans[u] = ans[v] + 1
q.append(u)
print(' '.join(str(x) for x in ans))
# Made By Mostafa_Khaled
```
| 94,522 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A subway scheme, classic for all Berland cities is represented by a set of n stations connected by n passages, each of which connects exactly two stations and does not pass through any others. Besides, in the classic scheme one can get from any station to any other one along the passages. The passages can be used to move in both directions. Between each pair of stations there is no more than one passage.
Berland mathematicians have recently proved a theorem that states that any classic scheme has a ringroad. There can be only one ringroad. In other words, in any classic scheme one can find the only scheme consisting of stations (where any two neighbouring ones are linked by a passage) and this cycle doesn't contain any station more than once.
This invention had a powerful social impact as now the stations could be compared according to their distance from the ringroad. For example, a citizen could say "I live in three passages from the ringroad" and another one could reply "you loser, I live in one passage from the ringroad". The Internet soon got filled with applications that promised to count the distance from the station to the ringroad (send a text message to a short number...).
The Berland government decided to put an end to these disturbances and start to control the situation. You are requested to write a program that can determine the remoteness from the ringroad for each station by the city subway scheme.
Input
The first line contains an integer n (3 β€ n β€ 3000), n is the number of stations (and trains at the same time) in the subway scheme. Then n lines contain descriptions of the trains, one per line. Each line contains a pair of integers xi, yi (1 β€ xi, yi β€ n) and represents the presence of a passage from station xi to station yi. The stations are numbered from 1 to n in an arbitrary order. It is guaranteed that xi β yi and that no pair of stations contain more than one passage. The passages can be used to travel both ways. It is guaranteed that the given description represents a classic subway scheme.
Output
Print n numbers. Separate the numbers by spaces, the i-th one should be equal to the distance of the i-th station from the ringroad. For the ringroad stations print number 0.
Examples
Input
4
1 3
4 3
4 2
1 2
Output
0 0 0 0
Input
6
1 2
3 4
6 4
2 3
1 3
3 5
Output
0 0 0 1 1 2
Tags: dfs and similar, graphs
Correct Solution:
```
n = int(input())
t, p = [0] * (n + 1), [[] for i in range(n + 1)]
for i in range(n):
a, b = map(int, input().split())
p[a].append(b)
p[b].append(a)
q, x, y = [1], 0, 0
while not x:
i = q.pop()
for j in p[i]:
if t[i] == j: continue
if t[j]:
t[j], x, y = i, j, t[j]
break
t[j] = i
q.append(j)
r = [y]
while x != y:
r.append(x)
x = t[x]
t = [-1] * (n + 1)
for i in r:
t[i] = 0
for k in r:
q = [j for j in p[k] if t[j]]
for i in q:
t[i] = 1
while q:
i = q.pop()
for j in p[i]:
if t[j] < 0:
t[j] = t[i] + 1
q.append(j)
print(' '.join(map(str, t[1:])))
```
| 94,523 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A subway scheme, classic for all Berland cities is represented by a set of n stations connected by n passages, each of which connects exactly two stations and does not pass through any others. Besides, in the classic scheme one can get from any station to any other one along the passages. The passages can be used to move in both directions. Between each pair of stations there is no more than one passage.
Berland mathematicians have recently proved a theorem that states that any classic scheme has a ringroad. There can be only one ringroad. In other words, in any classic scheme one can find the only scheme consisting of stations (where any two neighbouring ones are linked by a passage) and this cycle doesn't contain any station more than once.
This invention had a powerful social impact as now the stations could be compared according to their distance from the ringroad. For example, a citizen could say "I live in three passages from the ringroad" and another one could reply "you loser, I live in one passage from the ringroad". The Internet soon got filled with applications that promised to count the distance from the station to the ringroad (send a text message to a short number...).
The Berland government decided to put an end to these disturbances and start to control the situation. You are requested to write a program that can determine the remoteness from the ringroad for each station by the city subway scheme.
Input
The first line contains an integer n (3 β€ n β€ 3000), n is the number of stations (and trains at the same time) in the subway scheme. Then n lines contain descriptions of the trains, one per line. Each line contains a pair of integers xi, yi (1 β€ xi, yi β€ n) and represents the presence of a passage from station xi to station yi. The stations are numbered from 1 to n in an arbitrary order. It is guaranteed that xi β yi and that no pair of stations contain more than one passage. The passages can be used to travel both ways. It is guaranteed that the given description represents a classic subway scheme.
Output
Print n numbers. Separate the numbers by spaces, the i-th one should be equal to the distance of the i-th station from the ringroad. For the ringroad stations print number 0.
Examples
Input
4
1 3
4 3
4 2
1 2
Output
0 0 0 0
Input
6
1 2
3 4
6 4
2 3
1 3
3 5
Output
0 0 0 1 1 2
Tags: dfs and similar, graphs
Correct Solution:
```
from collections import deque
def findCycle(grafo):
n = len(grafo)
degree = dict()
for i in range(n):
degree[i] = len(grafo[i])
visited = [False for i in range(n)]
q = deque()
while True:
for i in range(len(degree)):
if degree[i] == 1 and visited[i] == False:
q.append(i)
if len(q) == 0:
break
while q:
temp = q.popleft()
visited[temp] = True
for destino in grafo[temp]:
degree[destino] -= 1
result = []
for i in range(len(visited)):
if visited[i] == False:
result.append(i)
return result
def distanciasAlCiclo(grafo, ciclo, numero_grande=float("inf")):
distancias = [numero_grande for i in range(n)]
for nodo in ciclo:
distancias[nodo] = 0
#padre = [-1]*n
cola = deque(ciclo)
while len(cola):
nodo = cola.popleft()
for conectado in grafo[nodo]:
if distancias[conectado] == numero_grande:
distancias[conectado] = distancias[nodo] + 1
#padre[conectado] = nodo
cola.append(conectado)
return distancias
"""
def distancia_a_nodos(grafo, nodo, nodos_destinos, padres=[]):
if nodo in nodos_destinos:
return 0
if len(grafo[nodo]) == 0:
return -1
for nodo in grafo[nodo]:
if nodo not in padres:
distancia = distancia_a_nodos(grafo, nodo, nodos_destinos, padres+[nodo])
if distancia != -1:
return distancia + 1
return -1
"""
n = int(input())
grafo = [list() for i in range(n)]
for i in range(n):
x, y = map(lambda z: int(z)-1, input().split())
grafo[x].append(y)
grafo[y].append(x)
nodos_ciclo = findCycle(grafo)
#for i in range(n):
# print(distancia_a_nodos(grafo, i, nodos_ciclo), end=" ")
#print()
print(" ".join(map(str, distanciasAlCiclo(grafo, nodos_ciclo))))
```
| 94,524 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A subway scheme, classic for all Berland cities is represented by a set of n stations connected by n passages, each of which connects exactly two stations and does not pass through any others. Besides, in the classic scheme one can get from any station to any other one along the passages. The passages can be used to move in both directions. Between each pair of stations there is no more than one passage.
Berland mathematicians have recently proved a theorem that states that any classic scheme has a ringroad. There can be only one ringroad. In other words, in any classic scheme one can find the only scheme consisting of stations (where any two neighbouring ones are linked by a passage) and this cycle doesn't contain any station more than once.
This invention had a powerful social impact as now the stations could be compared according to their distance from the ringroad. For example, a citizen could say "I live in three passages from the ringroad" and another one could reply "you loser, I live in one passage from the ringroad". The Internet soon got filled with applications that promised to count the distance from the station to the ringroad (send a text message to a short number...).
The Berland government decided to put an end to these disturbances and start to control the situation. You are requested to write a program that can determine the remoteness from the ringroad for each station by the city subway scheme.
Input
The first line contains an integer n (3 β€ n β€ 3000), n is the number of stations (and trains at the same time) in the subway scheme. Then n lines contain descriptions of the trains, one per line. Each line contains a pair of integers xi, yi (1 β€ xi, yi β€ n) and represents the presence of a passage from station xi to station yi. The stations are numbered from 1 to n in an arbitrary order. It is guaranteed that xi β yi and that no pair of stations contain more than one passage. The passages can be used to travel both ways. It is guaranteed that the given description represents a classic subway scheme.
Output
Print n numbers. Separate the numbers by spaces, the i-th one should be equal to the distance of the i-th station from the ringroad. For the ringroad stations print number 0.
Examples
Input
4
1 3
4 3
4 2
1 2
Output
0 0 0 0
Input
6
1 2
3 4
6 4
2 3
1 3
3 5
Output
0 0 0 1 1 2
Tags: dfs and similar, graphs
Correct Solution:
```
#!/usr/bin/pypy3
from sys import stdin,stderr
from collections import defaultdict
def readInts(): return map(int,stdin.readline().strip().split())
def print_err(*args,**kwargs): print(*args,file=stderr,**kwargs)
def find_cycle(n,g):
parents = [ None for _ in range(n+1) ]
parents[1] = 1
s = [1]
while s:
#print(s)
node = s.pop()
for nn in g[node]:
if nn==parents[node]: continue
if parents[nn] is not None:
#print(node,nn)
out = [nn]
lca = parents[nn]
while node:
out.append(node)
if node == lca: return out
node = parents[node]
parents[nn] = node
s.append(nn)
return None
def find_dists(n,g,ring):
dists = [ 0 for _ in range(n+1) ]
fringe = set(ring)
disc = set(ring)
d = 0
while fringe:
fringe_next = set()
for n in fringe:
dists[n] = d
for nn in g[n]:
if nn in disc: continue
fringe_next.add(nn)
disc = disc.union(fringe_next)
fringe = fringe_next
d += 1
return dists
def solve(n,g):
ring = find_cycle(n,g)
#print(ring)
dists = find_dists(n,g,ring)
return dists
def run():
n, = readInts()
g = defaultdict(list)
for _ in range(n):
a,b = readInts()
g[a].append(b)
g[b].append(a)
vs = solve(n,g)[1:]
print(" ".join(map(str,vs)))
run()
```
| 94,525 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A subway scheme, classic for all Berland cities is represented by a set of n stations connected by n passages, each of which connects exactly two stations and does not pass through any others. Besides, in the classic scheme one can get from any station to any other one along the passages. The passages can be used to move in both directions. Between each pair of stations there is no more than one passage.
Berland mathematicians have recently proved a theorem that states that any classic scheme has a ringroad. There can be only one ringroad. In other words, in any classic scheme one can find the only scheme consisting of stations (where any two neighbouring ones are linked by a passage) and this cycle doesn't contain any station more than once.
This invention had a powerful social impact as now the stations could be compared according to their distance from the ringroad. For example, a citizen could say "I live in three passages from the ringroad" and another one could reply "you loser, I live in one passage from the ringroad". The Internet soon got filled with applications that promised to count the distance from the station to the ringroad (send a text message to a short number...).
The Berland government decided to put an end to these disturbances and start to control the situation. You are requested to write a program that can determine the remoteness from the ringroad for each station by the city subway scheme.
Input
The first line contains an integer n (3 β€ n β€ 3000), n is the number of stations (and trains at the same time) in the subway scheme. Then n lines contain descriptions of the trains, one per line. Each line contains a pair of integers xi, yi (1 β€ xi, yi β€ n) and represents the presence of a passage from station xi to station yi. The stations are numbered from 1 to n in an arbitrary order. It is guaranteed that xi β yi and that no pair of stations contain more than one passage. The passages can be used to travel both ways. It is guaranteed that the given description represents a classic subway scheme.
Output
Print n numbers. Separate the numbers by spaces, the i-th one should be equal to the distance of the i-th station from the ringroad. For the ringroad stations print number 0.
Examples
Input
4
1 3
4 3
4 2
1 2
Output
0 0 0 0
Input
6
1 2
3 4
6 4
2 3
1 3
3 5
Output
0 0 0 1 1 2
Tags: dfs and similar, graphs
Correct Solution:
```
#!/usr/bin/pypy3
from sys import stdin,stderr
from collections import defaultdict
def readInts(): return map(int,stdin.readline().strip().split())
def print_err(*args,**kwargs): print(*args,file=stderr,**kwargs)
def find_cycle(g):
path = [None,1]
path_set = set(path)
s = [ (path,path_set) ]
while s:
#print(s)
path,path_set = s.pop()
node = path[-1]
parent = path[-2]
for nn in g[node]:
if nn==parent: continue
if nn in path_set:
out = []
while path:
out.append(path.pop())
if nn == out[-1]: return out
path_set2 = set(path_set)
path2 = list(path)
path_set2.add(nn)
path2.append(nn)
s.append( (path2,path_set2) )
return None
def find_dists(n,g,ring):
dists = [ 0 for _ in range(n+1) ]
fringe = set(ring)
disc = set(ring)
d = 0
while fringe:
fringe_next = set()
for n in fringe:
dists[n] = d
for nn in g[n]:
if nn in disc: continue
fringe_next.add(nn)
disc = disc.union(fringe_next)
fringe = fringe_next
d += 1
return dists
def solve(n,g):
ring = find_cycle(g)
#print(ring)
dists = find_dists(n,g,ring)
return dists
def run():
n, = readInts()
g = defaultdict(list)
for _ in range(n):
a,b = readInts()
g[a].append(b)
g[b].append(a)
vs = solve(n,g)[1:]
print(" ".join(map(str,vs)))
run()
```
| 94,526 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A subway scheme, classic for all Berland cities is represented by a set of n stations connected by n passages, each of which connects exactly two stations and does not pass through any others. Besides, in the classic scheme one can get from any station to any other one along the passages. The passages can be used to move in both directions. Between each pair of stations there is no more than one passage.
Berland mathematicians have recently proved a theorem that states that any classic scheme has a ringroad. There can be only one ringroad. In other words, in any classic scheme one can find the only scheme consisting of stations (where any two neighbouring ones are linked by a passage) and this cycle doesn't contain any station more than once.
This invention had a powerful social impact as now the stations could be compared according to their distance from the ringroad. For example, a citizen could say "I live in three passages from the ringroad" and another one could reply "you loser, I live in one passage from the ringroad". The Internet soon got filled with applications that promised to count the distance from the station to the ringroad (send a text message to a short number...).
The Berland government decided to put an end to these disturbances and start to control the situation. You are requested to write a program that can determine the remoteness from the ringroad for each station by the city subway scheme.
Input
The first line contains an integer n (3 β€ n β€ 3000), n is the number of stations (and trains at the same time) in the subway scheme. Then n lines contain descriptions of the trains, one per line. Each line contains a pair of integers xi, yi (1 β€ xi, yi β€ n) and represents the presence of a passage from station xi to station yi. The stations are numbered from 1 to n in an arbitrary order. It is guaranteed that xi β yi and that no pair of stations contain more than one passage. The passages can be used to travel both ways. It is guaranteed that the given description represents a classic subway scheme.
Output
Print n numbers. Separate the numbers by spaces, the i-th one should be equal to the distance of the i-th station from the ringroad. For the ringroad stations print number 0.
Examples
Input
4
1 3
4 3
4 2
1 2
Output
0 0 0 0
Input
6
1 2
3 4
6 4
2 3
1 3
3 5
Output
0 0 0 1 1 2
Tags: dfs and similar, graphs
Correct Solution:
```
from sys import setrecursionlimit
setrecursionlimit(30000)
def cycle(f):
v = f
cyc[v] = True
v = p[v]
while v != f:
cyc[v] = True
v = p[v]
def dfs(v, par):
global c, p, cycleroot
if c[v] == 2:
return
p[v] = par
if c[v] == 1:
cycle(v)
cycleroot = v
return True
c[v] = 1
for u in graph[v]:
if u != par:
if dfs(u, v):
return True
c[v] = 2
return False
def dfs1(v, d):
if used[v]:
return
used[v] = True
if cyc[v]:
dist[v] = 0
else:
dist[v] = d
for u in graph[v]:
dfs1(u, dist[v] + 1)
n = int(input())
m = n
graph = [[] for i in range(n)]
p = [-1] * n
c = [0] * n
cyc = [False] * n
if m != n:
print('NO')
exit(0)
for i in range(m):
a, b = map(int, input().split())
a, b = a - 1, b - 1
graph[a].append(b)
graph[b].append(a)
dfs(0, -1)
used = [False] * n
dist = [-1] * n
dfs1(cycleroot, 0)
print(' '.join([str(i) for i in dist]))
```
| 94,527 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A subway scheme, classic for all Berland cities is represented by a set of n stations connected by n passages, each of which connects exactly two stations and does not pass through any others. Besides, in the classic scheme one can get from any station to any other one along the passages. The passages can be used to move in both directions. Between each pair of stations there is no more than one passage.
Berland mathematicians have recently proved a theorem that states that any classic scheme has a ringroad. There can be only one ringroad. In other words, in any classic scheme one can find the only scheme consisting of stations (where any two neighbouring ones are linked by a passage) and this cycle doesn't contain any station more than once.
This invention had a powerful social impact as now the stations could be compared according to their distance from the ringroad. For example, a citizen could say "I live in three passages from the ringroad" and another one could reply "you loser, I live in one passage from the ringroad". The Internet soon got filled with applications that promised to count the distance from the station to the ringroad (send a text message to a short number...).
The Berland government decided to put an end to these disturbances and start to control the situation. You are requested to write a program that can determine the remoteness from the ringroad for each station by the city subway scheme.
Input
The first line contains an integer n (3 β€ n β€ 3000), n is the number of stations (and trains at the same time) in the subway scheme. Then n lines contain descriptions of the trains, one per line. Each line contains a pair of integers xi, yi (1 β€ xi, yi β€ n) and represents the presence of a passage from station xi to station yi. The stations are numbered from 1 to n in an arbitrary order. It is guaranteed that xi β yi and that no pair of stations contain more than one passage. The passages can be used to travel both ways. It is guaranteed that the given description represents a classic subway scheme.
Output
Print n numbers. Separate the numbers by spaces, the i-th one should be equal to the distance of the i-th station from the ringroad. For the ringroad stations print number 0.
Examples
Input
4
1 3
4 3
4 2
1 2
Output
0 0 0 0
Input
6
1 2
3 4
6 4
2 3
1 3
3 5
Output
0 0 0 1 1 2
Tags: dfs and similar, graphs
Correct Solution:
```
def dfs_non_recursive(graph, source):
path = []
stack = [source]
fathers = [x for x in range(len(graph))]
cycle = [0 for x in range(len(graph))]
while(len(stack) != 0):
v = stack.pop()
if v not in path:
path.append(v)
for w in range(len(graph)):
if graph[v][w]:
if fathers[v] == w:
continue
if w in path:
start = w
end = v
current = end
while current != start:
cycle[current] = 1
current = fathers[current]
cycle[current] = 1
return cycle
else:
fathers[w] = v
stack.append(w)
'''
for neighbor in graph[s]:
if neighbor in path:
return neighbor
stack.append(neighbor)
'''
return 0
def getDistances(graph, source, cycle):
stack = [source]
d = [-1 for x in range(len(graph))]
d[source] = 0
while len(stack) != 0:
v = stack.pop()
for w in range(len(graph)):
if graph[v][w] and d[w] == -1:
d[w] = d[v] + 1
stack.append(w)
for i in range(len(cycle)):
if cycle[i] == 1:
d[i] = 0
return d
def solve(graph):
n = len(graph)
fathers = [x for x in range(n)]
mark = [0 for x in range(n)]
#start = [-1]
#end = [-1]
#getInitCycle(graph, fathers, mark, 0, start, end)
cycle = dfs_non_recursive(graph,0)
temp = [0 if x == 1 else -1 for x in cycle]
temp.append(-1)
_graph = [[0 for x in range(n + 1)] for x in range(n + 1)]
for i in range(n):
for j in range(n):
if graph[i][j]:#Ambos vertices pertenecen al ciclo, se ovbian
if cycle[i] == 1 and cycle[j] == 1:
continue
elif cycle[i] == 1 and cycle[j] == 0:#Un vertice pertenece al ciclo
_graph[len(_graph) - 1][j] = 1
elif cycle[i] == 0 and cycle[j] == 1:#Un vertice pertenece al ciclo
_graph[i][len(_graph) -1] = 1
else: #NIngun vertice pertenece al ciclo
_graph[i][j] = 1
#temp[len(temp) - 1] = 0
mark = [0 for x in range(n + 1)]
d = getDistances(_graph, len(temp) - 1, cycle)#Llamar solamente con el nuevo vertice
result = []
for i in range(n):
result.append(d[i])
return result
#def getDistances(graph, mark, temp, v):
# mark[v] = 1
# for w in range(len(graph)):
# if graph[v][w] == 1 and mark[w] == 0:
# temp[w] = temp[v] + 1
# getDistances(graph, mark, temp, w)
def getInitCycle(graph, fathers, mark, v, start, end):#Devuelve donde inicia y termina el ciclo
mark[v] = 1
for w in range(len(graph)):
if graph[v][w] == 1:
if mark[w] == 0:
fathers[w] = v #Guardar quien lo llamo para obtener despues el ciclo
getInitCycle(graph, fathers, mark, w, start, end)
else:
if(fathers[v] != w and start[0] == -1): #El nodo ya esta visitado
start[0] = w #significa que existe un ciclo
end[0] = v
def getCycle(graph, fathers, start, end): #Devuelve el array cycle, donde se encuantra el ciclo
cycle = [0 for x in range(len(graph))]
current = end[0]
while current != start[0]:
cycle[current] = 1
current = fathers[current]
cycle[current] = 1
return cycle
n = int(input())
graph = [[0 for x in range(n)] for x in range(n)]
for item in range(n):
u,v = map(int, input().split())
graph[u - 1][v - 1] = 1
graph[v - 1][u - 1] = 1
result = solve(graph)
tmp = str(result[0])
for i in range(1, n):
tmp += " " + str(result[i])
print(tmp)
```
| 94,528 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A subway scheme, classic for all Berland cities is represented by a set of n stations connected by n passages, each of which connects exactly two stations and does not pass through any others. Besides, in the classic scheme one can get from any station to any other one along the passages. The passages can be used to move in both directions. Between each pair of stations there is no more than one passage.
Berland mathematicians have recently proved a theorem that states that any classic scheme has a ringroad. There can be only one ringroad. In other words, in any classic scheme one can find the only scheme consisting of stations (where any two neighbouring ones are linked by a passage) and this cycle doesn't contain any station more than once.
This invention had a powerful social impact as now the stations could be compared according to their distance from the ringroad. For example, a citizen could say "I live in three passages from the ringroad" and another one could reply "you loser, I live in one passage from the ringroad". The Internet soon got filled with applications that promised to count the distance from the station to the ringroad (send a text message to a short number...).
The Berland government decided to put an end to these disturbances and start to control the situation. You are requested to write a program that can determine the remoteness from the ringroad for each station by the city subway scheme.
Input
The first line contains an integer n (3 β€ n β€ 3000), n is the number of stations (and trains at the same time) in the subway scheme. Then n lines contain descriptions of the trains, one per line. Each line contains a pair of integers xi, yi (1 β€ xi, yi β€ n) and represents the presence of a passage from station xi to station yi. The stations are numbered from 1 to n in an arbitrary order. It is guaranteed that xi β yi and that no pair of stations contain more than one passage. The passages can be used to travel both ways. It is guaranteed that the given description represents a classic subway scheme.
Output
Print n numbers. Separate the numbers by spaces, the i-th one should be equal to the distance of the i-th station from the ringroad. For the ringroad stations print number 0.
Examples
Input
4
1 3
4 3
4 2
1 2
Output
0 0 0 0
Input
6
1 2
3 4
6 4
2 3
1 3
3 5
Output
0 0 0 1 1 2
Submitted Solution:
```
def dfs_non_recursive(graph, source):
n = len(graph)
stack = [source]
mark = [0 for x in range(n)]
fathers = [x for x in range(len(graph))]
cycle = [0 for x in range(len(graph))]
while len(stack) > 0:
v = stack.pop()
mark[v] = 1
for w in range(len(graph)):
if graph[v][w]:
if fathers[v] == w:
continue
if mark[w]:
start = w
end = v
current = end
while current != start:
cycle[current] = 1
current = fathers[current]
cycle[current] = 1
return cycle
else:
fathers[w] = v
stack.append(w)
def getDistances(graph, source, cycle):
stack = [source]
d = [-1 for x in range(len(graph))]
d[source] = 0
while len(stack) != 0:
v = stack.pop()
for w in range(len(graph)):
if graph[v][w] and d[w] == -1:
d[w] = d[v] + 1
stack.append(w)
for i in range(len(cycle)):
if cycle[i] == 1:
d[i] = 0
return d
def solve(graph):
n = len(graph)
cycle = dfs_non_recursive(graph,0)
_graph = [[0 for x in range(n + 1)] for x in range(n + 1)]
for i in range(n):
for j in range(n):
if graph[i][j]:#Ambos vertices pertenecen al ciclo, se ovbian
if cycle[i] == 1 and cycle[j] == 1:
continue
elif cycle[i] == 1 and cycle[j] == 0:#Un vertice pertenece al ciclo
_graph[len(_graph) - 1][j] = 1
elif cycle[i] == 0 and cycle[j] == 1:#Un vertice pertenece al ciclo
_graph[i][len(_graph) -1] = 1
else: #NIngun vertice pertenece al ciclo
_graph[i][j] = 1
d = getDistances(_graph, n, cycle)#Llamar solamente con el nuevo vertice
result = []
for i in range(n):
result.append(d[i])
return result
n = int(input())
graph = [[0 for x in range(n)] for x in range(n)]
for item in range(n):
u,v = map(int, input().split())
graph[u - 1][v - 1] = 1
graph[v - 1][u - 1] = 1
result = solve(graph)
tmp = str(result[0])
for i in range(1, n):
tmp += " " + str(result[i])
print(tmp)
```
Yes
| 94,529 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A subway scheme, classic for all Berland cities is represented by a set of n stations connected by n passages, each of which connects exactly two stations and does not pass through any others. Besides, in the classic scheme one can get from any station to any other one along the passages. The passages can be used to move in both directions. Between each pair of stations there is no more than one passage.
Berland mathematicians have recently proved a theorem that states that any classic scheme has a ringroad. There can be only one ringroad. In other words, in any classic scheme one can find the only scheme consisting of stations (where any two neighbouring ones are linked by a passage) and this cycle doesn't contain any station more than once.
This invention had a powerful social impact as now the stations could be compared according to their distance from the ringroad. For example, a citizen could say "I live in three passages from the ringroad" and another one could reply "you loser, I live in one passage from the ringroad". The Internet soon got filled with applications that promised to count the distance from the station to the ringroad (send a text message to a short number...).
The Berland government decided to put an end to these disturbances and start to control the situation. You are requested to write a program that can determine the remoteness from the ringroad for each station by the city subway scheme.
Input
The first line contains an integer n (3 β€ n β€ 3000), n is the number of stations (and trains at the same time) in the subway scheme. Then n lines contain descriptions of the trains, one per line. Each line contains a pair of integers xi, yi (1 β€ xi, yi β€ n) and represents the presence of a passage from station xi to station yi. The stations are numbered from 1 to n in an arbitrary order. It is guaranteed that xi β yi and that no pair of stations contain more than one passage. The passages can be used to travel both ways. It is guaranteed that the given description represents a classic subway scheme.
Output
Print n numbers. Separate the numbers by spaces, the i-th one should be equal to the distance of the i-th station from the ringroad. For the ringroad stations print number 0.
Examples
Input
4
1 3
4 3
4 2
1 2
Output
0 0 0 0
Input
6
1 2
3 4
6 4
2 3
1 3
3 5
Output
0 0 0 1 1 2
Submitted Solution:
```
#!/usr/bin/env python3
# -*- coding: utf-8 -*-
"""
Created on Wed Sep 2 14:44:47 2020
@author: suneelvarma
"""
from collections import defaultdict, deque
#edges = [(1,2), (3,4), (6,4), (2,3), (1,3), (3,5)]
#n= 6
#edges = [(1,2), (2,3), (3,4), (4,5), (1,5)]
#n= 5
'''
Approach:
Find nodes which are involved in a cycle.
Calculate distance from every node to any node involved in the graph.
Two Functions:
1) Return a couple of nodes involved in circle
2) Calculate distances
'''
def nodeInCycle(graph, n):
'''Returns a couple of nodes involved in circle'''
global time
nodesInCycle = []
arrivalTime = [0]+[float('inf') for _ in range(n)]
def dfs(node, parent):
global time
if arrivalTime[node] != float('inf'):
return arrivalTime[node]
arrivalTime[node] = time
minArrival = float('inf')
for child in graph.get(node,set()):
if child != parent:
time += 1
minArrival = min(dfs(child, node), minArrival)
if minArrival <= arrivalTime[node]:
nodesInCycle.append(node)
arrivalTime[node] = minArrival
return arrivalTime[node]
dfs(1, None)
return nodesInCycle
def distanceFromRing(graph, n):
''' return distances of each from the ring'''
nodesInCycle = nodeInCycle(graph, n)
distance = [0]+[float('inf') for _ in range(1,n+1)]
for node in nodesInCycle: distance[node] = 0
queue = deque([(node, 0) for node in nodesInCycle])
while queue:
node, distFromRing = queue.popleft()
for child in graph.get(node,set()):
if distance[child] == float('inf'):
distance[child] = distFromRing+1
queue.append((child, distance[child]))
return distance
def main(edges, n):
'''Main program, creates graph and calls for output'''
graph = defaultdict(set)
for u,v in edges:
graph[u].add(v)
graph[v].add(u)
return distanceFromRing(graph, n)[1:]
if __name__ == '__main__':
import sys
sys.setrecursionlimit(10000)
n = int(input())
edges = []
time = 1
for _ in range(n):
u,v = tuple(map(int,input().split()))
edges.append((u,v))
# time = 1
# n = 3000
# a = 1
# edges = []
# for b in range(2,3000):
# edges.append((a,b))
# a = b
#
# edges.append((2999,3000))
# edges.append((1,3000))
#edges = [(1,2), (3,4), (6,4), (2,3), (1,3), (3,5)]
#n= 6
print(*main(edges,n))
```
Yes
| 94,530 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A subway scheme, classic for all Berland cities is represented by a set of n stations connected by n passages, each of which connects exactly two stations and does not pass through any others. Besides, in the classic scheme one can get from any station to any other one along the passages. The passages can be used to move in both directions. Between each pair of stations there is no more than one passage.
Berland mathematicians have recently proved a theorem that states that any classic scheme has a ringroad. There can be only one ringroad. In other words, in any classic scheme one can find the only scheme consisting of stations (where any two neighbouring ones are linked by a passage) and this cycle doesn't contain any station more than once.
This invention had a powerful social impact as now the stations could be compared according to their distance from the ringroad. For example, a citizen could say "I live in three passages from the ringroad" and another one could reply "you loser, I live in one passage from the ringroad". The Internet soon got filled with applications that promised to count the distance from the station to the ringroad (send a text message to a short number...).
The Berland government decided to put an end to these disturbances and start to control the situation. You are requested to write a program that can determine the remoteness from the ringroad for each station by the city subway scheme.
Input
The first line contains an integer n (3 β€ n β€ 3000), n is the number of stations (and trains at the same time) in the subway scheme. Then n lines contain descriptions of the trains, one per line. Each line contains a pair of integers xi, yi (1 β€ xi, yi β€ n) and represents the presence of a passage from station xi to station yi. The stations are numbered from 1 to n in an arbitrary order. It is guaranteed that xi β yi and that no pair of stations contain more than one passage. The passages can be used to travel both ways. It is guaranteed that the given description represents a classic subway scheme.
Output
Print n numbers. Separate the numbers by spaces, the i-th one should be equal to the distance of the i-th station from the ringroad. For the ringroad stations print number 0.
Examples
Input
4
1 3
4 3
4 2
1 2
Output
0 0 0 0
Input
6
1 2
3 4
6 4
2 3
1 3
3 5
Output
0 0 0 1 1 2
Submitted Solution:
```
from collections import deque
import sys
sys.setrecursionlimit(5000)
class Graph(object):
def __init__(self, vertices, adj):
self.vertices = vertices#[v for v in range(1, vertices+1)]
self.dist = self.init_dist(vertices)
self.adj = adj
def init_dist(self, v):
d = {}
for i in range(1, v+1):
d[i] = 0
return d
# ~ def get_all_vertices(self):
# ~ return self.vertices
def iter_vertices(self):
for v in range(1,self.vertices+1):
yield v
def adj_list(self, v):
return self.adj[v]
class dfs_result:
def __init__(self):
self.dist = {}
self.parent = {}
self.finish_time = {}
self.start_time = {}
self.edges = {}
self.order = []
self.t = 0
self.loop = None
def dfs(g):
result = dfs_result()
for vertex in g.iter_vertices():
if vertex not in result.parent:
dfs_visit(g, vertex, result )
return result
def dfs_visit(g, v, result, parent = None):
result.parent[v] = parent
result.t += 1
result.start_time[v] = result.t
if parent:
result.edges[(parent, v)] = 'tree'
for n in g.adj_list(v):
if n not in result.parent:
dfs_visit(g, n, result, v)
elif (n not in result.finish_time):
if (n,v) not in result.edges:
result.edges[(v, n)] = 'back'
result.loop = [v, n]
elif result.start_time[v] < result.finish_time[n]:
result.edges[(v, n)] = 'forward'
else:
result.edges[(v, n)] = 'cross'
result.t += 1
result.finish_time[v] = result.t
result.order.append(v)
class BFSResult:
def __init__(self):
self.level = {}
self.parent = {}
def bfs(g, r, loop_dict, s):
r.parent = {s:None}
r.level = {s:0}
queue = deque()
queue.append(s)
while queue:
u = queue.popleft()
for n in g.adj_list(u):
if (n not in r.level) and n not in loop_dict:
r.parent[n] = u
r.level[n] = r.level[u] + 1
g.dist[n] = r.level[n]
queue.append(n)
if __name__ == '__main__':
t = int(input().rstrip())
adj = {}
for i in range(t):
u, v = list(map(int, input().rstrip().split()))
try:
adj[u].append(v)
except KeyError:
adj[u] = [v]
try:
adj[v].append(u)
except KeyError:
adj[v] = [u]
g = Graph(t, adj)
r = dfs(g)
# ~ print(r.loop)
# ~ print(r.edges)
start = r.loop[1]
end = r.loop[0]
loop = [end]
loop_dict = {end:1}
while r.parent[end]!=start:
loop.append(r.parent[end])
loop_dict[r.parent[end]] = 1
end = r.parent[end]
loop.append(start)
loop_dict[start] = 1
br = BFSResult()
for s in loop:
bfs(g, br, loop_dict, s)
d = g.dist
for k in d:
print(d[k], end = ' ')
```
Yes
| 94,531 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A subway scheme, classic for all Berland cities is represented by a set of n stations connected by n passages, each of which connects exactly two stations and does not pass through any others. Besides, in the classic scheme one can get from any station to any other one along the passages. The passages can be used to move in both directions. Between each pair of stations there is no more than one passage.
Berland mathematicians have recently proved a theorem that states that any classic scheme has a ringroad. There can be only one ringroad. In other words, in any classic scheme one can find the only scheme consisting of stations (where any two neighbouring ones are linked by a passage) and this cycle doesn't contain any station more than once.
This invention had a powerful social impact as now the stations could be compared according to their distance from the ringroad. For example, a citizen could say "I live in three passages from the ringroad" and another one could reply "you loser, I live in one passage from the ringroad". The Internet soon got filled with applications that promised to count the distance from the station to the ringroad (send a text message to a short number...).
The Berland government decided to put an end to these disturbances and start to control the situation. You are requested to write a program that can determine the remoteness from the ringroad for each station by the city subway scheme.
Input
The first line contains an integer n (3 β€ n β€ 3000), n is the number of stations (and trains at the same time) in the subway scheme. Then n lines contain descriptions of the trains, one per line. Each line contains a pair of integers xi, yi (1 β€ xi, yi β€ n) and represents the presence of a passage from station xi to station yi. The stations are numbered from 1 to n in an arbitrary order. It is guaranteed that xi β yi and that no pair of stations contain more than one passage. The passages can be used to travel both ways. It is guaranteed that the given description represents a classic subway scheme.
Output
Print n numbers. Separate the numbers by spaces, the i-th one should be equal to the distance of the i-th station from the ringroad. For the ringroad stations print number 0.
Examples
Input
4
1 3
4 3
4 2
1 2
Output
0 0 0 0
Input
6
1 2
3 4
6 4
2 3
1 3
3 5
Output
0 0 0 1 1 2
Submitted Solution:
```
from collections import *
read_line = lambda: [int(i) for i in input().split(' ')]
n = read_line()[0]
g, deg = [[] for i in range(n)], [0] * n
for i in range(n):
a, b = [v - 1 for v in read_line()]
g[a].append(b)
g[b].append(a)
deg[a] += 1
deg[b] += 1
ans = [0] * n
q = deque(v for v in range(n) if deg[v] == 1)
while q:
v = q.popleft()
ans[v] = None
for u in g[v]:
deg[u] -= 1
if deg[u] == 1:
q.append(u)
q = deque(v for v in range(n) if ans[v] == 0)
while q:
v = q.popleft()
for u in g[v]:
if ans[u] is None:
ans[u] = ans[v] + 1
q.append(u)
print(' '.join(str(x) for x in ans))
```
Yes
| 94,532 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A subway scheme, classic for all Berland cities is represented by a set of n stations connected by n passages, each of which connects exactly two stations and does not pass through any others. Besides, in the classic scheme one can get from any station to any other one along the passages. The passages can be used to move in both directions. Between each pair of stations there is no more than one passage.
Berland mathematicians have recently proved a theorem that states that any classic scheme has a ringroad. There can be only one ringroad. In other words, in any classic scheme one can find the only scheme consisting of stations (where any two neighbouring ones are linked by a passage) and this cycle doesn't contain any station more than once.
This invention had a powerful social impact as now the stations could be compared according to their distance from the ringroad. For example, a citizen could say "I live in three passages from the ringroad" and another one could reply "you loser, I live in one passage from the ringroad". The Internet soon got filled with applications that promised to count the distance from the station to the ringroad (send a text message to a short number...).
The Berland government decided to put an end to these disturbances and start to control the situation. You are requested to write a program that can determine the remoteness from the ringroad for each station by the city subway scheme.
Input
The first line contains an integer n (3 β€ n β€ 3000), n is the number of stations (and trains at the same time) in the subway scheme. Then n lines contain descriptions of the trains, one per line. Each line contains a pair of integers xi, yi (1 β€ xi, yi β€ n) and represents the presence of a passage from station xi to station yi. The stations are numbered from 1 to n in an arbitrary order. It is guaranteed that xi β yi and that no pair of stations contain more than one passage. The passages can be used to travel both ways. It is guaranteed that the given description represents a classic subway scheme.
Output
Print n numbers. Separate the numbers by spaces, the i-th one should be equal to the distance of the i-th station from the ringroad. For the ringroad stations print number 0.
Examples
Input
4
1 3
4 3
4 2
1 2
Output
0 0 0 0
Input
6
1 2
3 4
6 4
2 3
1 3
3 5
Output
0 0 0 1 1 2
Submitted Solution:
```
def dfs_non_recursive(graph, source):
path = []
stack = [source]
fathers = [x for x in range(len(graph))]
cycle = [0 for x in range(len(graph))]
while(len(stack) != 0):
v = stack.pop()
if v not in path:
path.append(v)
for w in range(len(graph)):
if graph[v][w]:
if fathers[v] == w:
continue
if w in path:
start = w
end = v
current = end
while current != start:
cycle[current] = 1
current = fathers[current]
cycle[current] = 1
return cycle
else:
fathers[w] = v
stack.append(w)
return 0
def solve(graph):
n = len(graph)
fathers = [x for x in range(n)]
mark = [0 for x in range(n)]
#start = [-1]
#end = [-1]
#getInitCycle(graph, fathers, mark, 0, start, end)
cycle = dfs_non_recursive(graph, 0)
result = [0 for x in range(n)]
for i in range(n):
if cycle[i] == 1:
result = getDistances(graph, i, cycle)#Llamar al metodo
#solamente en los vertices del ciclo
return result
def getDistances(graph, source, cycle):
stack = [source]
d = [-1 for x in range(len(graph))]
d[source] = 0
for i in range(len(cycle)):
if cycle[i] == 1:
d[i] = 0
while len(stack) != 0:
v = stack.pop()
for w in range(len(graph)):
if graph[v][w] and d[w] == -1:
d[w] = d[v] + 1
stack.append(w)
return d
n = int(input())
graph = [[0 for x in range(n)] for x in range(n)]
for item in range(n):
u,v = map(int, input().split())
graph[u - 1][v - 1] = 1
graph[v - 1][u - 1] = 1
result = solve(graph)
tmp = str(result[0])
for i in range(1, n):
tmp += " " + str(result[i])
print(tmp)
```
No
| 94,533 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A subway scheme, classic for all Berland cities is represented by a set of n stations connected by n passages, each of which connects exactly two stations and does not pass through any others. Besides, in the classic scheme one can get from any station to any other one along the passages. The passages can be used to move in both directions. Between each pair of stations there is no more than one passage.
Berland mathematicians have recently proved a theorem that states that any classic scheme has a ringroad. There can be only one ringroad. In other words, in any classic scheme one can find the only scheme consisting of stations (where any two neighbouring ones are linked by a passage) and this cycle doesn't contain any station more than once.
This invention had a powerful social impact as now the stations could be compared according to their distance from the ringroad. For example, a citizen could say "I live in three passages from the ringroad" and another one could reply "you loser, I live in one passage from the ringroad". The Internet soon got filled with applications that promised to count the distance from the station to the ringroad (send a text message to a short number...).
The Berland government decided to put an end to these disturbances and start to control the situation. You are requested to write a program that can determine the remoteness from the ringroad for each station by the city subway scheme.
Input
The first line contains an integer n (3 β€ n β€ 3000), n is the number of stations (and trains at the same time) in the subway scheme. Then n lines contain descriptions of the trains, one per line. Each line contains a pair of integers xi, yi (1 β€ xi, yi β€ n) and represents the presence of a passage from station xi to station yi. The stations are numbered from 1 to n in an arbitrary order. It is guaranteed that xi β yi and that no pair of stations contain more than one passage. The passages can be used to travel both ways. It is guaranteed that the given description represents a classic subway scheme.
Output
Print n numbers. Separate the numbers by spaces, the i-th one should be equal to the distance of the i-th station from the ringroad. For the ringroad stations print number 0.
Examples
Input
4
1 3
4 3
4 2
1 2
Output
0 0 0 0
Input
6
1 2
3 4
6 4
2 3
1 3
3 5
Output
0 0 0 1 1 2
Submitted Solution:
```
def dfs_non_recursive(graph, source):
path = []
stack = [source]
fathers = [x for x in range(len(graph))]
cycle = [0 for x in range(len(graph))]
while(len(stack) != 0):
v = stack.pop()
if v not in path:
path.append(v)
for w in range(len(graph)):
if graph[v][w]:
if fathers[v] == w:
continue
if w in path:
start = w
end = v
current = end
while current != start:
cycle[current] = 1
current = fathers[current]
cycle[current] = 1
return cycle
else:
fathers[w] = v
stack.append(w)
return 0
def solve(graph):
n = len(graph)
fathers = [x for x in range(n)]
mark = [0 for x in range(n)]
cycle = dfs_non_recursive(graph, 0)
print(cycle)
result = [0 for x in range(n)]
for i in range(n):
if cycle[i] == 1:
tmp = getDistances(graph, i, cycle)#Llamar al metodo
#solamente en los vertices del ciclo
for i in range(n):
if tmp[i] != -1:
result[i] = tmp[i]
return result
def getDistances(graph, source, cycle):
stack = [source]
d = [-1 for x in range(len(graph))]
d[source] = 0
for i in range(len(cycle)):
if cycle[i] == 1:
d[i] = 0
while len(stack) != 0:
v = stack.pop()
for w in range(len(graph)):
if graph[v][w] and d[w] == -1:
d[w] = d[v] + 1
stack.append(w)
return d
n = int(input())
graph = [[0 for x in range(n)] for x in range(n)]
for item in range(n):
u,v = map(int, input().split())
graph[u - 1][v - 1] = 1
graph[v - 1][u - 1] = 1
result = solve(graph)
tmp = str(result[0])
for i in range(1, n):
tmp += " " + str(result[i])
print(tmp)
'''
5
2 5
4 5
4 3
4 2
1 4
'''
```
No
| 94,534 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A subway scheme, classic for all Berland cities is represented by a set of n stations connected by n passages, each of which connects exactly two stations and does not pass through any others. Besides, in the classic scheme one can get from any station to any other one along the passages. The passages can be used to move in both directions. Between each pair of stations there is no more than one passage.
Berland mathematicians have recently proved a theorem that states that any classic scheme has a ringroad. There can be only one ringroad. In other words, in any classic scheme one can find the only scheme consisting of stations (where any two neighbouring ones are linked by a passage) and this cycle doesn't contain any station more than once.
This invention had a powerful social impact as now the stations could be compared according to their distance from the ringroad. For example, a citizen could say "I live in three passages from the ringroad" and another one could reply "you loser, I live in one passage from the ringroad". The Internet soon got filled with applications that promised to count the distance from the station to the ringroad (send a text message to a short number...).
The Berland government decided to put an end to these disturbances and start to control the situation. You are requested to write a program that can determine the remoteness from the ringroad for each station by the city subway scheme.
Input
The first line contains an integer n (3 β€ n β€ 3000), n is the number of stations (and trains at the same time) in the subway scheme. Then n lines contain descriptions of the trains, one per line. Each line contains a pair of integers xi, yi (1 β€ xi, yi β€ n) and represents the presence of a passage from station xi to station yi. The stations are numbered from 1 to n in an arbitrary order. It is guaranteed that xi β yi and that no pair of stations contain more than one passage. The passages can be used to travel both ways. It is guaranteed that the given description represents a classic subway scheme.
Output
Print n numbers. Separate the numbers by spaces, the i-th one should be equal to the distance of the i-th station from the ringroad. For the ringroad stations print number 0.
Examples
Input
4
1 3
4 3
4 2
1 2
Output
0 0 0 0
Input
6
1 2
3 4
6 4
2 3
1 3
3 5
Output
0 0 0 1 1 2
Submitted Solution:
```
def ringroad(anterior,actual, recorrido, fin=False):
if fin:
return recorrido
interseccion = adj[actual].intersection(recorrido)
if len(interseccion) >= 2:
inicio = min(interseccion - {anterior})
recorrido = recorrido[recorrido.index(inicio):]
global RR
RR = set(recorrido)
return ringroad(anterior,actual,recorrido,True)
for estacion in adj[actual]:
if estacion != anterior:
if estacion not in recorrido:
recorrido.append(estacion)
ringroad(actual, estacion, recorrido)
else:
continue
def distanciaRR(i):
minimo = 100000
global adj
global RR
if len(RR.intersection(adj[i])) >= 1:
return 1
else:
for v in adj[i]:
dist = 1 + distanciaRR(v)
if dist < minimo:
minimo = dist
return minimo
adj = dict()
RR = set()
#Ingreso de datos
n = int(input())
for i in range(n):
x,y = list(map(int,input().split()))
if x in adj.keys():
adj[x].add(y)
else:
adj[x] = {y}
if y in adj.keys():
adj[y].add(x)
else:
adj[y] = {x}
#Procesado
ringroad(-1,1,[1])
out = ""
for i in range(1,n+1):
if i in RR:
out += "0 "
else:
out += str(distanciaRR(i)) + " "
print(out)
```
No
| 94,535 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A subway scheme, classic for all Berland cities is represented by a set of n stations connected by n passages, each of which connects exactly two stations and does not pass through any others. Besides, in the classic scheme one can get from any station to any other one along the passages. The passages can be used to move in both directions. Between each pair of stations there is no more than one passage.
Berland mathematicians have recently proved a theorem that states that any classic scheme has a ringroad. There can be only one ringroad. In other words, in any classic scheme one can find the only scheme consisting of stations (where any two neighbouring ones are linked by a passage) and this cycle doesn't contain any station more than once.
This invention had a powerful social impact as now the stations could be compared according to their distance from the ringroad. For example, a citizen could say "I live in three passages from the ringroad" and another one could reply "you loser, I live in one passage from the ringroad". The Internet soon got filled with applications that promised to count the distance from the station to the ringroad (send a text message to a short number...).
The Berland government decided to put an end to these disturbances and start to control the situation. You are requested to write a program that can determine the remoteness from the ringroad for each station by the city subway scheme.
Input
The first line contains an integer n (3 β€ n β€ 3000), n is the number of stations (and trains at the same time) in the subway scheme. Then n lines contain descriptions of the trains, one per line. Each line contains a pair of integers xi, yi (1 β€ xi, yi β€ n) and represents the presence of a passage from station xi to station yi. The stations are numbered from 1 to n in an arbitrary order. It is guaranteed that xi β yi and that no pair of stations contain more than one passage. The passages can be used to travel both ways. It is guaranteed that the given description represents a classic subway scheme.
Output
Print n numbers. Separate the numbers by spaces, the i-th one should be equal to the distance of the i-th station from the ringroad. For the ringroad stations print number 0.
Examples
Input
4
1 3
4 3
4 2
1 2
Output
0 0 0 0
Input
6
1 2
3 4
6 4
2 3
1 3
3 5
Output
0 0 0 1 1 2
Submitted Solution:
```
from collections import deque
"""
def bfs(grafo, nodo_inicial, numero_grande=float("inf")):
ciclo = numero_grande
distancias = [numero_grande]*n
distancias[nodo_inicial] = 0
padre = [-1]*n
cola = deque([nodo_inicial])
while len(cola):
nodo = cola.popleft()
for conectado in grafo[nodo]:
if distancias[conectado] == numero_grande:
distancias[conectado] = distancias[nodo] + 1
padre[conectado] = nodo
cola.append(conectado)
elif padre[nodo] != conectado and padre[conectado] != nodo:
ciclo = min(ciclo, distancias[nodo] + distancias[conectado] + 1)
print(padre, nodo, conectado)
print(padre[padre[conectado]])
print(padre[padre[padre[conectado]]])
return
"""
def bfs(grafo, nodo_inicial, V, visited, numero_grande=float("inf")):
ciclo = numero_grande
distancias = [numero_grande]*V
distancias[nodo_inicial] = 0
padre = [-1] * V
cola = deque([nodo_inicial])
visited[nodo_inicial] = True
first = True
while len(cola):
nodo = cola.popleft()
#if visited[nodo] and not first:
# print(visited)
# return 42
#if first:
# first = False
for conectado in grafo[nodo]:
if distancias[conectado] == numero_grande:
distancias[conectado] = distancias[nodo] + 1
padre[conectado] = nodo
cola.append(conectado)
visited[conectado] = True
elif padre[nodo] != conectado and padre[conectado] != nodo:
ciclo = min(ciclo, distancias[nodo] + distancias[conectado] + 1)
#print(padre, nodo, conectado)
#print(padre[padre[conectado]])
#print(padre[padre[padre[conectado]]])
return ciclo, distancias, padre
"""
if not visited[v]:
visited[v] = True
q.append(v)
parent[v] = u
elif parent[u] != v:
return True
"""
return ciclo, distancias, padre
def shortest_cycle(grafo, numero_grande=float("inf")):
ciclo = numero_grande
nodos_mas_cortos = [False] * n
for i in range(n):
visitados = [False] * n
cantidad_nodos, distancias, padres = bfs(grafo, i, n, visitados, numero_grande=5000)
if cantidad_nodos < ciclo:
#print(distancias)
#print(padres)
ciclo = cantidad_nodos
nodos_mas_cortos = list(visitados)
if cantidad_nodos == 3:
break
return nodos_mas_cortos, distancias, padres
"""
# https://stackoverflow.com/a/41936867/6292472
def bfs(grafo, nodo_inicial, V, visited, numero_grande=float("inf")):
nodo_padre = nodo_inicial
cola = deque([nodo_inicial])
while len(cola):
nodo = cola.popleft()
if visited[nodo]:
print(visited)
return -2
visited[nodo] = True
for conectado in grafo[nodo]:
if conectado != nodo_padre:
cola.append(conectado)
# cola.extend(grafo[nodo])
nodo_padre = nodo
return 0
"""
def encontrar_distancia_al_ciclo(grafo, nodo_inicial, nodos_del_ciclo):
if nodo_inicial in nodos_del_ciclo:
return 0
if len(grafo[nodo_inicial]) == 0:
return -1
if len(grafo[nodo_inicial]) == 1:
distancia = encontrar_distancia_al_ciclo(grafo, grafo[nodo_inicial][0], nodos_del_ciclo)
if distancia == -1:
return -1
return distancia + 1
for nodo in grafo[nodo_inicial]:
distancia = encontrar_distancia_al_ciclo(grafo, grafo[nodo_inicial][0], nodos_del_ciclo)
if distancia != -1:
return distancia + 1
return -1
n = int(input())
grafo = [list() for i in range(n)]
for i in range(n):
x, y = map(lambda z: int(z)-1, input().split())
grafo[x].append(y)
grafo[y].append(x)
#print(shortest_cycle(grafo))
nodos_mas_cortos, distancias, padres = shortest_cycle(grafo)
#print(distancias)
#print(padres)
#for i in distancias:
# print(max(i-1, 0), end=" ")
#print()
#print(nodos_mas_cortos)
##print(distancias)
#print(padres)
"""
for i in range(len(nodos_mas_cortos)):
if nodos_mas_cortos[i]:
pass
#print("0 ", end="")
else:
print(padres[i])
"""
"""
nodos_del_ciclo = []
for nodo in padres:
if padres[nodo] <= 0:
nodos_del_ciclo.append(nodo)
"""
nodos_del_ciclo = []
for nodo in range(n):
if nodos_mas_cortos[nodo]:
nodos_del_ciclo.append(nodo)
#print(nodos_del_ciclo)
#print(nodos_mas_cortos)
#print(grafo)
for i in range(n):
if i in nodos_del_ciclo:
pass
print(0, end=" ")
else:
#distancia = 0
#encontrado = False
#print(i, grafo[i])
#print(i, encontrar_distancia_al_ciclo(grafo, i, nodos_del_ciclo))
distancia = encontrar_distancia_al_ciclo(grafo, i, nodos_del_ciclo)
"""
mi_padre = padres[i]
while not encontrado:
distancia += 1
if mi_padre in nodos_del_ciclo:
encontrado = True
mi_padre = padres[mi_padre]
#mi_padre = padres[i]
#print(mi_padre in nodos_del_ciclo)
#if mi_padre not in nodos_del_ciclo:
# print("\t", padres[mi_padre] in nodos_del_ciclo)
#print(grafo[padres[i]])
#while True:
"""
print(distancia, end=" ")
print()
#print(bfs(grafo, n-1, n, visitados, numero_grande=5000))
#print(visitados)
#visitados = [False] * n
#print(bfs(grafo, n-1, n, visitados, numero_grande=5000))
#print(visitados)
```
No
| 94,536 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You have integer n. Calculate how many ways are there to fully cover belt-like area of 4n-2 triangles with diamond shapes.
Diamond shape consists of two triangles. You can move, rotate or flip the shape, but you cannot scale it.
2 coverings are different if some 2 triangles are covered by the same diamond shape in one of them and by different diamond shapes in the other one.
Please look at pictures below for better understanding.
<image> On the left you can see the diamond shape you will use, and on the right you can see the area you want to fill.
<image> These are the figures of the area you want to fill for n = 1, 2, 3, 4.
You have to answer t independent test cases.
Input
The first line contains a single integer t (1 β€ t β€ 10^{4}) β the number of test cases.
Each of the next t lines contains a single integer n (1 β€ n β€ 10^{9}).
Output
For each test case, print the number of ways to fully cover belt-like area of 4n-2 triangles using diamond shape. It can be shown that under given constraints this number of ways doesn't exceed 10^{18}.
Example
Input
2
2
1
Output
2
1
Note
In the first test case, there are the following 2 ways to fill the area:
<image>
In the second test case, there is a unique way to fill the area:
<image>
Tags: brute force, dp, implementation, math
Correct Solution:
```
t = int(input())
while t > 0:
n = input()
print(n)
t -= 1
```
| 94,537 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You have integer n. Calculate how many ways are there to fully cover belt-like area of 4n-2 triangles with diamond shapes.
Diamond shape consists of two triangles. You can move, rotate or flip the shape, but you cannot scale it.
2 coverings are different if some 2 triangles are covered by the same diamond shape in one of them and by different diamond shapes in the other one.
Please look at pictures below for better understanding.
<image> On the left you can see the diamond shape you will use, and on the right you can see the area you want to fill.
<image> These are the figures of the area you want to fill for n = 1, 2, 3, 4.
You have to answer t independent test cases.
Input
The first line contains a single integer t (1 β€ t β€ 10^{4}) β the number of test cases.
Each of the next t lines contains a single integer n (1 β€ n β€ 10^{9}).
Output
For each test case, print the number of ways to fully cover belt-like area of 4n-2 triangles using diamond shape. It can be shown that under given constraints this number of ways doesn't exceed 10^{18}.
Example
Input
2
2
1
Output
2
1
Note
In the first test case, there are the following 2 ways to fill the area:
<image>
In the second test case, there is a unique way to fill the area:
<image>
Tags: brute force, dp, implementation, math
Correct Solution:
```
for nt in range(int(input())):
n=int(input())
print (n)
```
| 94,538 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You have integer n. Calculate how many ways are there to fully cover belt-like area of 4n-2 triangles with diamond shapes.
Diamond shape consists of two triangles. You can move, rotate or flip the shape, but you cannot scale it.
2 coverings are different if some 2 triangles are covered by the same diamond shape in one of them and by different diamond shapes in the other one.
Please look at pictures below for better understanding.
<image> On the left you can see the diamond shape you will use, and on the right you can see the area you want to fill.
<image> These are the figures of the area you want to fill for n = 1, 2, 3, 4.
You have to answer t independent test cases.
Input
The first line contains a single integer t (1 β€ t β€ 10^{4}) β the number of test cases.
Each of the next t lines contains a single integer n (1 β€ n β€ 10^{9}).
Output
For each test case, print the number of ways to fully cover belt-like area of 4n-2 triangles using diamond shape. It can be shown that under given constraints this number of ways doesn't exceed 10^{18}.
Example
Input
2
2
1
Output
2
1
Note
In the first test case, there are the following 2 ways to fill the area:
<image>
In the second test case, there is a unique way to fill the area:
<image>
Tags: brute force, dp, implementation, math
Correct Solution:
```
T = int(input())
for i in range(T):
st=input()
print(st)
```
| 94,539 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You have integer n. Calculate how many ways are there to fully cover belt-like area of 4n-2 triangles with diamond shapes.
Diamond shape consists of two triangles. You can move, rotate or flip the shape, but you cannot scale it.
2 coverings are different if some 2 triangles are covered by the same diamond shape in one of them and by different diamond shapes in the other one.
Please look at pictures below for better understanding.
<image> On the left you can see the diamond shape you will use, and on the right you can see the area you want to fill.
<image> These are the figures of the area you want to fill for n = 1, 2, 3, 4.
You have to answer t independent test cases.
Input
The first line contains a single integer t (1 β€ t β€ 10^{4}) β the number of test cases.
Each of the next t lines contains a single integer n (1 β€ n β€ 10^{9}).
Output
For each test case, print the number of ways to fully cover belt-like area of 4n-2 triangles using diamond shape. It can be shown that under given constraints this number of ways doesn't exceed 10^{18}.
Example
Input
2
2
1
Output
2
1
Note
In the first test case, there are the following 2 ways to fill the area:
<image>
In the second test case, there is a unique way to fill the area:
<image>
Tags: brute force, dp, implementation, math
Correct Solution:
```
# import sys
# sys.stdin = open('input.txt', 'r')
# sys.stdout = open('output.txt', 'w+')
for _ in range (0, int(input())):
n = int(input())
ans = 1
print(n)
```
| 94,540 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You have integer n. Calculate how many ways are there to fully cover belt-like area of 4n-2 triangles with diamond shapes.
Diamond shape consists of two triangles. You can move, rotate or flip the shape, but you cannot scale it.
2 coverings are different if some 2 triangles are covered by the same diamond shape in one of them and by different diamond shapes in the other one.
Please look at pictures below for better understanding.
<image> On the left you can see the diamond shape you will use, and on the right you can see the area you want to fill.
<image> These are the figures of the area you want to fill for n = 1, 2, 3, 4.
You have to answer t independent test cases.
Input
The first line contains a single integer t (1 β€ t β€ 10^{4}) β the number of test cases.
Each of the next t lines contains a single integer n (1 β€ n β€ 10^{9}).
Output
For each test case, print the number of ways to fully cover belt-like area of 4n-2 triangles using diamond shape. It can be shown that under given constraints this number of ways doesn't exceed 10^{18}.
Example
Input
2
2
1
Output
2
1
Note
In the first test case, there are the following 2 ways to fill the area:
<image>
In the second test case, there is a unique way to fill the area:
<image>
Tags: brute force, dp, implementation, math
Correct Solution:
```
def sol():
return int(input())
for _ in range(int(input())):
print(sol())
```
| 94,541 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You have integer n. Calculate how many ways are there to fully cover belt-like area of 4n-2 triangles with diamond shapes.
Diamond shape consists of two triangles. You can move, rotate or flip the shape, but you cannot scale it.
2 coverings are different if some 2 triangles are covered by the same diamond shape in one of them and by different diamond shapes in the other one.
Please look at pictures below for better understanding.
<image> On the left you can see the diamond shape you will use, and on the right you can see the area you want to fill.
<image> These are the figures of the area you want to fill for n = 1, 2, 3, 4.
You have to answer t independent test cases.
Input
The first line contains a single integer t (1 β€ t β€ 10^{4}) β the number of test cases.
Each of the next t lines contains a single integer n (1 β€ n β€ 10^{9}).
Output
For each test case, print the number of ways to fully cover belt-like area of 4n-2 triangles using diamond shape. It can be shown that under given constraints this number of ways doesn't exceed 10^{18}.
Example
Input
2
2
1
Output
2
1
Note
In the first test case, there are the following 2 ways to fill the area:
<image>
In the second test case, there is a unique way to fill the area:
<image>
Tags: brute force, dp, implementation, math
Correct Solution:
```
a = int(input())
for i in range (a):
print(int(input()))
```
| 94,542 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You have integer n. Calculate how many ways are there to fully cover belt-like area of 4n-2 triangles with diamond shapes.
Diamond shape consists of two triangles. You can move, rotate or flip the shape, but you cannot scale it.
2 coverings are different if some 2 triangles are covered by the same diamond shape in one of them and by different diamond shapes in the other one.
Please look at pictures below for better understanding.
<image> On the left you can see the diamond shape you will use, and on the right you can see the area you want to fill.
<image> These are the figures of the area you want to fill for n = 1, 2, 3, 4.
You have to answer t independent test cases.
Input
The first line contains a single integer t (1 β€ t β€ 10^{4}) β the number of test cases.
Each of the next t lines contains a single integer n (1 β€ n β€ 10^{9}).
Output
For each test case, print the number of ways to fully cover belt-like area of 4n-2 triangles using diamond shape. It can be shown that under given constraints this number of ways doesn't exceed 10^{18}.
Example
Input
2
2
1
Output
2
1
Note
In the first test case, there are the following 2 ways to fill the area:
<image>
In the second test case, there is a unique way to fill the area:
<image>
Tags: brute force, dp, implementation, math
Correct Solution:
```
from sys import stdin
def main():
r = stdin.readline
cases = int(r())
for case in range(cases):
n = int(r())
print(n)
main()
```
| 94,543 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You have integer n. Calculate how many ways are there to fully cover belt-like area of 4n-2 triangles with diamond shapes.
Diamond shape consists of two triangles. You can move, rotate or flip the shape, but you cannot scale it.
2 coverings are different if some 2 triangles are covered by the same diamond shape in one of them and by different diamond shapes in the other one.
Please look at pictures below for better understanding.
<image> On the left you can see the diamond shape you will use, and on the right you can see the area you want to fill.
<image> These are the figures of the area you want to fill for n = 1, 2, 3, 4.
You have to answer t independent test cases.
Input
The first line contains a single integer t (1 β€ t β€ 10^{4}) β the number of test cases.
Each of the next t lines contains a single integer n (1 β€ n β€ 10^{9}).
Output
For each test case, print the number of ways to fully cover belt-like area of 4n-2 triangles using diamond shape. It can be shown that under given constraints this number of ways doesn't exceed 10^{18}.
Example
Input
2
2
1
Output
2
1
Note
In the first test case, there are the following 2 ways to fill the area:
<image>
In the second test case, there is a unique way to fill the area:
<image>
Tags: brute force, dp, implementation, math
Correct Solution:
```
t = int(input())
b = []
for i in range(t):
n = int(input())
b.append(n)
for i in range(t):
print(b[i])
```
| 94,544 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have integer n. Calculate how many ways are there to fully cover belt-like area of 4n-2 triangles with diamond shapes.
Diamond shape consists of two triangles. You can move, rotate or flip the shape, but you cannot scale it.
2 coverings are different if some 2 triangles are covered by the same diamond shape in one of them and by different diamond shapes in the other one.
Please look at pictures below for better understanding.
<image> On the left you can see the diamond shape you will use, and on the right you can see the area you want to fill.
<image> These are the figures of the area you want to fill for n = 1, 2, 3, 4.
You have to answer t independent test cases.
Input
The first line contains a single integer t (1 β€ t β€ 10^{4}) β the number of test cases.
Each of the next t lines contains a single integer n (1 β€ n β€ 10^{9}).
Output
For each test case, print the number of ways to fully cover belt-like area of 4n-2 triangles using diamond shape. It can be shown that under given constraints this number of ways doesn't exceed 10^{18}.
Example
Input
2
2
1
Output
2
1
Note
In the first test case, there are the following 2 ways to fill the area:
<image>
In the second test case, there is a unique way to fill the area:
<image>
Submitted Solution:
```
t = int(input())
for i in range(t):
k = int(input())
print(k)
```
Yes
| 94,545 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have integer n. Calculate how many ways are there to fully cover belt-like area of 4n-2 triangles with diamond shapes.
Diamond shape consists of two triangles. You can move, rotate or flip the shape, but you cannot scale it.
2 coverings are different if some 2 triangles are covered by the same diamond shape in one of them and by different diamond shapes in the other one.
Please look at pictures below for better understanding.
<image> On the left you can see the diamond shape you will use, and on the right you can see the area you want to fill.
<image> These are the figures of the area you want to fill for n = 1, 2, 3, 4.
You have to answer t independent test cases.
Input
The first line contains a single integer t (1 β€ t β€ 10^{4}) β the number of test cases.
Each of the next t lines contains a single integer n (1 β€ n β€ 10^{9}).
Output
For each test case, print the number of ways to fully cover belt-like area of 4n-2 triangles using diamond shape. It can be shown that under given constraints this number of ways doesn't exceed 10^{18}.
Example
Input
2
2
1
Output
2
1
Note
In the first test case, there are the following 2 ways to fill the area:
<image>
In the second test case, there is a unique way to fill the area:
<image>
Submitted Solution:
```
# t = int(input())
for _ in range(int(input())):
n = int(input())
print(n)
```
Yes
| 94,546 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have integer n. Calculate how many ways are there to fully cover belt-like area of 4n-2 triangles with diamond shapes.
Diamond shape consists of two triangles. You can move, rotate or flip the shape, but you cannot scale it.
2 coverings are different if some 2 triangles are covered by the same diamond shape in one of them and by different diamond shapes in the other one.
Please look at pictures below for better understanding.
<image> On the left you can see the diamond shape you will use, and on the right you can see the area you want to fill.
<image> These are the figures of the area you want to fill for n = 1, 2, 3, 4.
You have to answer t independent test cases.
Input
The first line contains a single integer t (1 β€ t β€ 10^{4}) β the number of test cases.
Each of the next t lines contains a single integer n (1 β€ n β€ 10^{9}).
Output
For each test case, print the number of ways to fully cover belt-like area of 4n-2 triangles using diamond shape. It can be shown that under given constraints this number of ways doesn't exceed 10^{18}.
Example
Input
2
2
1
Output
2
1
Note
In the first test case, there are the following 2 ways to fill the area:
<image>
In the second test case, there is a unique way to fill the area:
<image>
Submitted Solution:
```
t = int(input())
x = []
for i in range(t):
n = int(input())
x += [n]
for k in x:
print(k)
```
Yes
| 94,547 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have integer n. Calculate how many ways are there to fully cover belt-like area of 4n-2 triangles with diamond shapes.
Diamond shape consists of two triangles. You can move, rotate or flip the shape, but you cannot scale it.
2 coverings are different if some 2 triangles are covered by the same diamond shape in one of them and by different diamond shapes in the other one.
Please look at pictures below for better understanding.
<image> On the left you can see the diamond shape you will use, and on the right you can see the area you want to fill.
<image> These are the figures of the area you want to fill for n = 1, 2, 3, 4.
You have to answer t independent test cases.
Input
The first line contains a single integer t (1 β€ t β€ 10^{4}) β the number of test cases.
Each of the next t lines contains a single integer n (1 β€ n β€ 10^{9}).
Output
For each test case, print the number of ways to fully cover belt-like area of 4n-2 triangles using diamond shape. It can be shown that under given constraints this number of ways doesn't exceed 10^{18}.
Example
Input
2
2
1
Output
2
1
Note
In the first test case, there are the following 2 ways to fill the area:
<image>
In the second test case, there is a unique way to fill the area:
<image>
Submitted Solution:
```
t=int(input())
for _ in range(t):
valid=True
n=int(input())
print(n)
```
Yes
| 94,548 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have integer n. Calculate how many ways are there to fully cover belt-like area of 4n-2 triangles with diamond shapes.
Diamond shape consists of two triangles. You can move, rotate or flip the shape, but you cannot scale it.
2 coverings are different if some 2 triangles are covered by the same diamond shape in one of them and by different diamond shapes in the other one.
Please look at pictures below for better understanding.
<image> On the left you can see the diamond shape you will use, and on the right you can see the area you want to fill.
<image> These are the figures of the area you want to fill for n = 1, 2, 3, 4.
You have to answer t independent test cases.
Input
The first line contains a single integer t (1 β€ t β€ 10^{4}) β the number of test cases.
Each of the next t lines contains a single integer n (1 β€ n β€ 10^{9}).
Output
For each test case, print the number of ways to fully cover belt-like area of 4n-2 triangles using diamond shape. It can be shown that under given constraints this number of ways doesn't exceed 10^{18}.
Example
Input
2
2
1
Output
2
1
Note
In the first test case, there are the following 2 ways to fill the area:
<image>
In the second test case, there is a unique way to fill the area:
<image>
Submitted Solution:
```
for _ in range(int(input())):
n = int(input())
if(n==1):
print(1)
else:
print(2)
```
No
| 94,549 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have integer n. Calculate how many ways are there to fully cover belt-like area of 4n-2 triangles with diamond shapes.
Diamond shape consists of two triangles. You can move, rotate or flip the shape, but you cannot scale it.
2 coverings are different if some 2 triangles are covered by the same diamond shape in one of them and by different diamond shapes in the other one.
Please look at pictures below for better understanding.
<image> On the left you can see the diamond shape you will use, and on the right you can see the area you want to fill.
<image> These are the figures of the area you want to fill for n = 1, 2, 3, 4.
You have to answer t independent test cases.
Input
The first line contains a single integer t (1 β€ t β€ 10^{4}) β the number of test cases.
Each of the next t lines contains a single integer n (1 β€ n β€ 10^{9}).
Output
For each test case, print the number of ways to fully cover belt-like area of 4n-2 triangles using diamond shape. It can be shown that under given constraints this number of ways doesn't exceed 10^{18}.
Example
Input
2
2
1
Output
2
1
Note
In the first test case, there are the following 2 ways to fill the area:
<image>
In the second test case, there is a unique way to fill the area:
<image>
Submitted Solution:
```
from sys import stdin,stdout
t=int(stdin.readline())
for query in range(t):
n=int(stdin.readline())
if n==1:
stdout.write('1')
elif n==2:
stdout.write('2')
elif n==3:
stdout.write('3')
else:
stdout.write('4')
stdout.write('\n')
```
No
| 94,550 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have integer n. Calculate how many ways are there to fully cover belt-like area of 4n-2 triangles with diamond shapes.
Diamond shape consists of two triangles. You can move, rotate or flip the shape, but you cannot scale it.
2 coverings are different if some 2 triangles are covered by the same diamond shape in one of them and by different diamond shapes in the other one.
Please look at pictures below for better understanding.
<image> On the left you can see the diamond shape you will use, and on the right you can see the area you want to fill.
<image> These are the figures of the area you want to fill for n = 1, 2, 3, 4.
You have to answer t independent test cases.
Input
The first line contains a single integer t (1 β€ t β€ 10^{4}) β the number of test cases.
Each of the next t lines contains a single integer n (1 β€ n β€ 10^{9}).
Output
For each test case, print the number of ways to fully cover belt-like area of 4n-2 triangles using diamond shape. It can be shown that under given constraints this number of ways doesn't exceed 10^{18}.
Example
Input
2
2
1
Output
2
1
Note
In the first test case, there are the following 2 ways to fill the area:
<image>
In the second test case, there is a unique way to fill the area:
<image>
Submitted Solution:
```
cases = int(input())
for i in range(cases):
n = int(input())
r = int((4*n - 4)/2)
total = 1
while (r >= 1):
total = 1*r
r -= 1
print (total)
```
No
| 94,551 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have integer n. Calculate how many ways are there to fully cover belt-like area of 4n-2 triangles with diamond shapes.
Diamond shape consists of two triangles. You can move, rotate or flip the shape, but you cannot scale it.
2 coverings are different if some 2 triangles are covered by the same diamond shape in one of them and by different diamond shapes in the other one.
Please look at pictures below for better understanding.
<image> On the left you can see the diamond shape you will use, and on the right you can see the area you want to fill.
<image> These are the figures of the area you want to fill for n = 1, 2, 3, 4.
You have to answer t independent test cases.
Input
The first line contains a single integer t (1 β€ t β€ 10^{4}) β the number of test cases.
Each of the next t lines contains a single integer n (1 β€ n β€ 10^{9}).
Output
For each test case, print the number of ways to fully cover belt-like area of 4n-2 triangles using diamond shape. It can be shown that under given constraints this number of ways doesn't exceed 10^{18}.
Example
Input
2
2
1
Output
2
1
Note
In the first test case, there are the following 2 ways to fill the area:
<image>
In the second test case, there is a unique way to fill the area:
<image>
Submitted Solution:
```
t = int(input())
for _ in range(t):
n = int(input())
shape = 4*n-2
if n == 1:
print(1)
else:
print(2)
```
No
| 94,552 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Polygon is not only the best platform for developing problems but also a square matrix with side n, initially filled with the character 0.
On the polygon, military training was held. The soldiers placed a cannon above each cell in the first row and a cannon to the left of each cell in the first column. Thus, exactly 2n cannons were placed.
<image> Initial polygon for n=4.
Cannons shoot character 1. At any moment of time, no more than one cannon is shooting. When a 1 flies out of a cannon, it flies forward (in the direction of the shot) until it collides with a polygon border or another 1. After that, it takes the cell in which it was before the collision and remains there. Take a look at the examples for better understanding.
More formally:
* if a cannon stands in the row i, to the left of the first column, and shoots with a 1, then the 1 starts its flight from the cell (i, 1) and ends in some cell (i, j);
* if a cannon stands in the column j, above the first row, and shoots with a 1, then the 1 starts its flight from the cell (1, j) and ends in some cell (i, j).
For example, consider the following sequence of shots:
<image>
1. Shoot the cannon in the row 2. 2. Shoot the cannon in the row 2. 3. Shoot the cannon in column 3.
You have a report from the military training on your desk. This report is a square matrix with side length n consisting of 0 and 1. You wonder if the training actually happened. In other words, is there a sequence of shots such that, after the training, you get the given matrix?
Each cannon can make an arbitrary number of shots. Before the training, each cell of the polygon contains 0.
Input
The first line contains an integer t (1 β€ t β€ 1000) β the number of test cases. Then t test cases follow.
Each test case starts with a line containing an integer n (1 β€ n β€ 50) β the size of the polygon.
This is followed by n lines of length n, consisting of 0 and 1 β the polygon matrix after the training.
The total area of the matrices in all test cases in one test does not exceed 10^5.
Output
For each test case print:
* YES if there is a sequence of shots leading to a given matrix;
* NO if such a sequence does not exist.
The letters in the words YES and NO can be printed in any case.
Example
Input
5
4
0010
0011
0000
0000
2
10
01
2
00
00
4
0101
1111
0101
0111
4
0100
1110
0101
0111
Output
YES
NO
YES
YES
NO
Note
The first test case was explained in the statement.
The answer to the second test case is NO, since a 1 in a cell (1, 1) flying out of any cannon would continue its flight further.
Tags: dp, graphs, implementation, shortest paths
Correct Solution:
```
import math,string,itertools,fractions,heapq,collections,re,array,bisect,sys,copy,functools
# import time,random,resource
sys.setrecursionlimit(10**7)
inf = 10**20
eps = 1.0 / 10**10
mod = 10**9+7
mod2 = 998244353
dd = [(-1,0),(0,1),(1,0),(0,-1)]
ddn = [(-1,0),(-1,1),(0,1),(1,1),(1,0),(1,-1),(0,-1),(-1,-1)]
def LI(): return list(map(int, sys.stdin.readline().split()))
def LLI(): return [list(map(int, l.split())) for l in sys.stdin.readlines()]
def LI_(): return [int(x)-1 for x in sys.stdin.readline().split()]
def LF(): return [float(x) for x in sys.stdin.readline().split()]
def LS(): return sys.stdin.readline().split()
def I(): return int(sys.stdin.readline())
def F(): return float(sys.stdin.readline())
def S(): return input()
def pf(s): return print(s, flush=True)
def pe(s): return print(str(s), file=sys.stderr)
def JA(a, sep): return sep.join(map(str, a))
def JAA(a, s, t): return s.join(t.join(map(str, b)) for b in a)
def IF(c, t, f): return t if c else f
def main():
t = I()
rr = []
for _ in range(t):
n = I()
a = [S() for _ in range(n)]
ok = True
for i in range(n-1):
for j in range(n-1):
if a[i][j] == '1' and a[i][j+1] == '0' and a[i+1][j] == '0':
ok = False
rr.append(IF(ok, 'YES', 'NO'))
return JA(rr, "\n")
print(main())
```
| 94,553 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Polygon is not only the best platform for developing problems but also a square matrix with side n, initially filled with the character 0.
On the polygon, military training was held. The soldiers placed a cannon above each cell in the first row and a cannon to the left of each cell in the first column. Thus, exactly 2n cannons were placed.
<image> Initial polygon for n=4.
Cannons shoot character 1. At any moment of time, no more than one cannon is shooting. When a 1 flies out of a cannon, it flies forward (in the direction of the shot) until it collides with a polygon border or another 1. After that, it takes the cell in which it was before the collision and remains there. Take a look at the examples for better understanding.
More formally:
* if a cannon stands in the row i, to the left of the first column, and shoots with a 1, then the 1 starts its flight from the cell (i, 1) and ends in some cell (i, j);
* if a cannon stands in the column j, above the first row, and shoots with a 1, then the 1 starts its flight from the cell (1, j) and ends in some cell (i, j).
For example, consider the following sequence of shots:
<image>
1. Shoot the cannon in the row 2. 2. Shoot the cannon in the row 2. 3. Shoot the cannon in column 3.
You have a report from the military training on your desk. This report is a square matrix with side length n consisting of 0 and 1. You wonder if the training actually happened. In other words, is there a sequence of shots such that, after the training, you get the given matrix?
Each cannon can make an arbitrary number of shots. Before the training, each cell of the polygon contains 0.
Input
The first line contains an integer t (1 β€ t β€ 1000) β the number of test cases. Then t test cases follow.
Each test case starts with a line containing an integer n (1 β€ n β€ 50) β the size of the polygon.
This is followed by n lines of length n, consisting of 0 and 1 β the polygon matrix after the training.
The total area of the matrices in all test cases in one test does not exceed 10^5.
Output
For each test case print:
* YES if there is a sequence of shots leading to a given matrix;
* NO if such a sequence does not exist.
The letters in the words YES and NO can be printed in any case.
Example
Input
5
4
0010
0011
0000
0000
2
10
01
2
00
00
4
0101
1111
0101
0111
4
0100
1110
0101
0111
Output
YES
NO
YES
YES
NO
Note
The first test case was explained in the statement.
The answer to the second test case is NO, since a 1 in a cell (1, 1) flying out of any cannon would continue its flight further.
Tags: dp, graphs, implementation, shortest paths
Correct Solution:
```
import os
import sys
import io
import math
#input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline
for t in range(int(input())):
n=int(input())
matrix=[]
for i in range(n):
matrix.append(list(map(int,input())))
check=True
for y in range(n-1):
for x in range(n-1):
if matrix[y][x]==1:
if matrix[y][x+1]==0 and matrix[y+1][x]==0:
check=False
break
if check==True:
print("YES")
else:
print("NO")
```
| 94,554 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Polygon is not only the best platform for developing problems but also a square matrix with side n, initially filled with the character 0.
On the polygon, military training was held. The soldiers placed a cannon above each cell in the first row and a cannon to the left of each cell in the first column. Thus, exactly 2n cannons were placed.
<image> Initial polygon for n=4.
Cannons shoot character 1. At any moment of time, no more than one cannon is shooting. When a 1 flies out of a cannon, it flies forward (in the direction of the shot) until it collides with a polygon border or another 1. After that, it takes the cell in which it was before the collision and remains there. Take a look at the examples for better understanding.
More formally:
* if a cannon stands in the row i, to the left of the first column, and shoots with a 1, then the 1 starts its flight from the cell (i, 1) and ends in some cell (i, j);
* if a cannon stands in the column j, above the first row, and shoots with a 1, then the 1 starts its flight from the cell (1, j) and ends in some cell (i, j).
For example, consider the following sequence of shots:
<image>
1. Shoot the cannon in the row 2. 2. Shoot the cannon in the row 2. 3. Shoot the cannon in column 3.
You have a report from the military training on your desk. This report is a square matrix with side length n consisting of 0 and 1. You wonder if the training actually happened. In other words, is there a sequence of shots such that, after the training, you get the given matrix?
Each cannon can make an arbitrary number of shots. Before the training, each cell of the polygon contains 0.
Input
The first line contains an integer t (1 β€ t β€ 1000) β the number of test cases. Then t test cases follow.
Each test case starts with a line containing an integer n (1 β€ n β€ 50) β the size of the polygon.
This is followed by n lines of length n, consisting of 0 and 1 β the polygon matrix after the training.
The total area of the matrices in all test cases in one test does not exceed 10^5.
Output
For each test case print:
* YES if there is a sequence of shots leading to a given matrix;
* NO if such a sequence does not exist.
The letters in the words YES and NO can be printed in any case.
Example
Input
5
4
0010
0011
0000
0000
2
10
01
2
00
00
4
0101
1111
0101
0111
4
0100
1110
0101
0111
Output
YES
NO
YES
YES
NO
Note
The first test case was explained in the statement.
The answer to the second test case is NO, since a 1 in a cell (1, 1) flying out of any cannon would continue its flight further.
Tags: dp, graphs, implementation, shortest paths
Correct Solution:
```
t = int(input())
for _ in range(t):
n = int(input())
lines = []
for _ in range(n):
lines.append(input())
breaker = False
yas = True
for x in range(0,n-1):
if breaker:
break
else:
for y in range(0,n-1):
if lines[x][y] == '0' or (lines[x][y] == '1' and (lines[x+1][y] == '1' or lines[x][y+1] == '1')):
pass
else:
print("NO")
breaker = True
yas = False
break
if yas:
print("YES")
```
| 94,555 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Polygon is not only the best platform for developing problems but also a square matrix with side n, initially filled with the character 0.
On the polygon, military training was held. The soldiers placed a cannon above each cell in the first row and a cannon to the left of each cell in the first column. Thus, exactly 2n cannons were placed.
<image> Initial polygon for n=4.
Cannons shoot character 1. At any moment of time, no more than one cannon is shooting. When a 1 flies out of a cannon, it flies forward (in the direction of the shot) until it collides with a polygon border or another 1. After that, it takes the cell in which it was before the collision and remains there. Take a look at the examples for better understanding.
More formally:
* if a cannon stands in the row i, to the left of the first column, and shoots with a 1, then the 1 starts its flight from the cell (i, 1) and ends in some cell (i, j);
* if a cannon stands in the column j, above the first row, and shoots with a 1, then the 1 starts its flight from the cell (1, j) and ends in some cell (i, j).
For example, consider the following sequence of shots:
<image>
1. Shoot the cannon in the row 2. 2. Shoot the cannon in the row 2. 3. Shoot the cannon in column 3.
You have a report from the military training on your desk. This report is a square matrix with side length n consisting of 0 and 1. You wonder if the training actually happened. In other words, is there a sequence of shots such that, after the training, you get the given matrix?
Each cannon can make an arbitrary number of shots. Before the training, each cell of the polygon contains 0.
Input
The first line contains an integer t (1 β€ t β€ 1000) β the number of test cases. Then t test cases follow.
Each test case starts with a line containing an integer n (1 β€ n β€ 50) β the size of the polygon.
This is followed by n lines of length n, consisting of 0 and 1 β the polygon matrix after the training.
The total area of the matrices in all test cases in one test does not exceed 10^5.
Output
For each test case print:
* YES if there is a sequence of shots leading to a given matrix;
* NO if such a sequence does not exist.
The letters in the words YES and NO can be printed in any case.
Example
Input
5
4
0010
0011
0000
0000
2
10
01
2
00
00
4
0101
1111
0101
0111
4
0100
1110
0101
0111
Output
YES
NO
YES
YES
NO
Note
The first test case was explained in the statement.
The answer to the second test case is NO, since a 1 in a cell (1, 1) flying out of any cannon would continue its flight further.
Tags: dp, graphs, implementation, shortest paths
Correct Solution:
```
# 11111111111111111111111111111111111111111111111111
# 11111111111111111111111111111111111111111111111111
# 11111111111111111111111111111111111111111111111111
# 11111111111111111111111111111111111111111111111111
# 11111111111111111111111111111111111111111111111111
# 11111111111111111111111111111111111111111111111111
# 11111111111111111111111111111111111111111111111111
# 11111111111111111111111111111111111111111111111111
# 11111111111111111111111111111111111111111111111111
# 11111111111111111111111111111111111111111111111111
# 11111111111111111111111111111111111111111111111111
# 11111111111111111111111111111111111111111111111111
# 11111111111111111111111111111111111111111111111111
# 11111111111111111111111111111111111111111111111111
# 11111111111111111111111111111111111111111111111111
# 11111111111111111111111111111111111111111111111111
# 11111111111111111111111111111111111111111111111111
# 11111111111111111111111111111111111111111111111111
# 11111111111111111111111111111111111111111111111111
# 11111111111111111111111111111111111111111111111111
# 11111111111111111111111111111111111111111111111111
# 11111111111111111111111111111111111111111111111111
# 11111111111111111111111111111111111111111111111111
# 11111111111111111111111111111111111111111111111111
# 11111111111111111111111111111111111111111111111111
# 11111111111111111111111111111111111111111111111111
# 11111111111111111111111111111111111111111111111111
# 11111111111111111111111111111111111111111111111111
# 11111111111111111111111111111111111111111111111111
# 11111111111111111111111111111111111111111111111111
# 11111111111111111111111111111111111111111111111111
# 11111111111111111111111111111111111111111111111111
# 11111111111111111111111111111111111111111111111111
# 11111111111111111111111111111111111111111111111111
# 11111111111111111111111111111111111111111111111111
# 11111111111111111111111111111111111111111111111111
# 11111111111111111111111111111111111111111111111111
# 11111111111111111111111111111111111111111111111111
# 11111111111111111111111111111111111111111111111111
# 11111111111111111111111111111111111111111111111111
# 11111111111111111111111111111111111111111111111111
# 11111111111111111111111111111111111111111111111111
# 11111111111111111111111111111111111111111111111111
# 11111111111111111111111111111111111111111111111111
# 11111111111111111111111111111111111111111111111111
# 11111111111111111111111111111111111111111111111111
# 11111111111111111111111111111111111111111111111111
# 11111111111111111111111111111111111111111111111111
# 11111111111111111111111111111111111111111111111111
# 11111111111111111111111111111111111111111111111111
class Node:
def __init__(self, i, j, state, n) -> None:
self.i = i
self.j = j
self.state = state
self.childs = set()
self.is_border = i+1 == n or j+1 == n
def __repr__(self) -> str:
return f"Node(i={self.i}; j={self.j}; state={self.state})"
def solution():
n = int(input())
nodes = {
(i, j): Node(i, j, name, n)
for i in range(n)
for j, name in enumerate(input())
}
# for i1, i2 in zip(range(n), range(n-1)):
# for j1, j2 in zip(range(n-1), range(n)):
# nodes[(i1, j1)].childs.add(nodes[(i1, j1+1)])
# nodes[(i2, j2)].childs.add(nodes[(i2+1, j2)])
for i in range(n):
for j in range(n-1):
nodes[(i, j)].childs.add(nodes[(i, j+1)])
for i in range(n-1):
for j in range(n):
nodes[(i, j)].childs.add(nodes[(i+1, j)])
nodes = nodes.values()
for node in filter(lambda n: n.state == '1', nodes):
if can_travel(node):
continue
else:
print("NO")
break
else:
print("YES")
# for key in nodes:
# print(f"{key}:{nodes[key].childs}")
def can_travel(node: Node, visited={}):
if node.state == '0':
return False
elif node.state == '1':
if node.is_border:
return True
childs = list(filter(lambda child: child.state == '1', node.childs))
if len(childs) == 0:
return False
can = False
for child in childs:
if child not in visited:
visited[child] = can_travel(child, visited)
can = visited[child]
return can
else:
raise RuntimeError("undefined state")
if __name__ == "__main__":
#TODO: input
t = int(input())
for _ in range(t):
solution()
```
| 94,556 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Polygon is not only the best platform for developing problems but also a square matrix with side n, initially filled with the character 0.
On the polygon, military training was held. The soldiers placed a cannon above each cell in the first row and a cannon to the left of each cell in the first column. Thus, exactly 2n cannons were placed.
<image> Initial polygon for n=4.
Cannons shoot character 1. At any moment of time, no more than one cannon is shooting. When a 1 flies out of a cannon, it flies forward (in the direction of the shot) until it collides with a polygon border or another 1. After that, it takes the cell in which it was before the collision and remains there. Take a look at the examples for better understanding.
More formally:
* if a cannon stands in the row i, to the left of the first column, and shoots with a 1, then the 1 starts its flight from the cell (i, 1) and ends in some cell (i, j);
* if a cannon stands in the column j, above the first row, and shoots with a 1, then the 1 starts its flight from the cell (1, j) and ends in some cell (i, j).
For example, consider the following sequence of shots:
<image>
1. Shoot the cannon in the row 2. 2. Shoot the cannon in the row 2. 3. Shoot the cannon in column 3.
You have a report from the military training on your desk. This report is a square matrix with side length n consisting of 0 and 1. You wonder if the training actually happened. In other words, is there a sequence of shots such that, after the training, you get the given matrix?
Each cannon can make an arbitrary number of shots. Before the training, each cell of the polygon contains 0.
Input
The first line contains an integer t (1 β€ t β€ 1000) β the number of test cases. Then t test cases follow.
Each test case starts with a line containing an integer n (1 β€ n β€ 50) β the size of the polygon.
This is followed by n lines of length n, consisting of 0 and 1 β the polygon matrix after the training.
The total area of the matrices in all test cases in one test does not exceed 10^5.
Output
For each test case print:
* YES if there is a sequence of shots leading to a given matrix;
* NO if such a sequence does not exist.
The letters in the words YES and NO can be printed in any case.
Example
Input
5
4
0010
0011
0000
0000
2
10
01
2
00
00
4
0101
1111
0101
0111
4
0100
1110
0101
0111
Output
YES
NO
YES
YES
NO
Note
The first test case was explained in the statement.
The answer to the second test case is NO, since a 1 in a cell (1, 1) flying out of any cannon would continue its flight further.
Tags: dp, graphs, implementation, shortest paths
Correct Solution:
```
from collections import deque
T = int(input())
for _ in range(T):
n = int(input())
arr = []
for _ in range(n):
arr.append(list(map(int,list(input()))))
vis = [[False for _ in range(n)] for _ in range(n)]
q = deque()
for i in range(n):
if arr[n-1][i] == 1:
q.append((n-1,i))
if arr[i][n-1] == 1:
q.append((i,n-1))
while len(q)>0:
i,j = q.popleft()
if i<0 or j<0 or arr[i][j] != 1 or vis[i][j]:
continue
vis[i][j] = True
q.append((i-1,j))
q.append((i,j-1))
ans = "YES"
for i in range(n):
for j in range(n):
if arr[i][j] == 1 and not vis[i][j]:
ans = "NO"
print(ans)
```
| 94,557 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Polygon is not only the best platform for developing problems but also a square matrix with side n, initially filled with the character 0.
On the polygon, military training was held. The soldiers placed a cannon above each cell in the first row and a cannon to the left of each cell in the first column. Thus, exactly 2n cannons were placed.
<image> Initial polygon for n=4.
Cannons shoot character 1. At any moment of time, no more than one cannon is shooting. When a 1 flies out of a cannon, it flies forward (in the direction of the shot) until it collides with a polygon border or another 1. After that, it takes the cell in which it was before the collision and remains there. Take a look at the examples for better understanding.
More formally:
* if a cannon stands in the row i, to the left of the first column, and shoots with a 1, then the 1 starts its flight from the cell (i, 1) and ends in some cell (i, j);
* if a cannon stands in the column j, above the first row, and shoots with a 1, then the 1 starts its flight from the cell (1, j) and ends in some cell (i, j).
For example, consider the following sequence of shots:
<image>
1. Shoot the cannon in the row 2. 2. Shoot the cannon in the row 2. 3. Shoot the cannon in column 3.
You have a report from the military training on your desk. This report is a square matrix with side length n consisting of 0 and 1. You wonder if the training actually happened. In other words, is there a sequence of shots such that, after the training, you get the given matrix?
Each cannon can make an arbitrary number of shots. Before the training, each cell of the polygon contains 0.
Input
The first line contains an integer t (1 β€ t β€ 1000) β the number of test cases. Then t test cases follow.
Each test case starts with a line containing an integer n (1 β€ n β€ 50) β the size of the polygon.
This is followed by n lines of length n, consisting of 0 and 1 β the polygon matrix after the training.
The total area of the matrices in all test cases in one test does not exceed 10^5.
Output
For each test case print:
* YES if there is a sequence of shots leading to a given matrix;
* NO if such a sequence does not exist.
The letters in the words YES and NO can be printed in any case.
Example
Input
5
4
0010
0011
0000
0000
2
10
01
2
00
00
4
0101
1111
0101
0111
4
0100
1110
0101
0111
Output
YES
NO
YES
YES
NO
Note
The first test case was explained in the statement.
The answer to the second test case is NO, since a 1 in a cell (1, 1) flying out of any cannon would continue its flight further.
Tags: dp, graphs, implementation, shortest paths
Correct Solution:
```
from math import *
from collections import *
from operator import itemgetter
import bisect
i = lambda: input()
ii = lambda: int(input())
iia = lambda: list(map(int,input().split()))
isa = lambda: list(input().split())
I = lambda:list(map(int,input().split()))
chrIdx = lambda x: ord(x)-96
idxChr = lambda x: chr(96+x)
t = ii()
for _ in range(t):
n = ii()
m = [0]*n
for i in range(n):
m[i] = input()
flag = 1
for i in range(n):
for j in range(n):
if(m[i][j]=='1'):
if(i<n-1 and m[i+1][j]!='1') and (j<n-1 and m[i][j+1]!='1'):
#print(i,j)
flag = 0
break
if(flag==0):
break
if flag==0:
print("NO")
else:
print("YES")
```
| 94,558 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Polygon is not only the best platform for developing problems but also a square matrix with side n, initially filled with the character 0.
On the polygon, military training was held. The soldiers placed a cannon above each cell in the first row and a cannon to the left of each cell in the first column. Thus, exactly 2n cannons were placed.
<image> Initial polygon for n=4.
Cannons shoot character 1. At any moment of time, no more than one cannon is shooting. When a 1 flies out of a cannon, it flies forward (in the direction of the shot) until it collides with a polygon border or another 1. After that, it takes the cell in which it was before the collision and remains there. Take a look at the examples for better understanding.
More formally:
* if a cannon stands in the row i, to the left of the first column, and shoots with a 1, then the 1 starts its flight from the cell (i, 1) and ends in some cell (i, j);
* if a cannon stands in the column j, above the first row, and shoots with a 1, then the 1 starts its flight from the cell (1, j) and ends in some cell (i, j).
For example, consider the following sequence of shots:
<image>
1. Shoot the cannon in the row 2. 2. Shoot the cannon in the row 2. 3. Shoot the cannon in column 3.
You have a report from the military training on your desk. This report is a square matrix with side length n consisting of 0 and 1. You wonder if the training actually happened. In other words, is there a sequence of shots such that, after the training, you get the given matrix?
Each cannon can make an arbitrary number of shots. Before the training, each cell of the polygon contains 0.
Input
The first line contains an integer t (1 β€ t β€ 1000) β the number of test cases. Then t test cases follow.
Each test case starts with a line containing an integer n (1 β€ n β€ 50) β the size of the polygon.
This is followed by n lines of length n, consisting of 0 and 1 β the polygon matrix after the training.
The total area of the matrices in all test cases in one test does not exceed 10^5.
Output
For each test case print:
* YES if there is a sequence of shots leading to a given matrix;
* NO if such a sequence does not exist.
The letters in the words YES and NO can be printed in any case.
Example
Input
5
4
0010
0011
0000
0000
2
10
01
2
00
00
4
0101
1111
0101
0111
4
0100
1110
0101
0111
Output
YES
NO
YES
YES
NO
Note
The first test case was explained in the statement.
The answer to the second test case is NO, since a 1 in a cell (1, 1) flying out of any cannon would continue its flight further.
Tags: dp, graphs, implementation, shortest paths
Correct Solution:
```
t = int(input().strip())
for _ in range(t):
g = []
flag = 0
n = int(input().strip())
for _ in range(n):
g.append(input().strip())
for i in range(n - 1):
for j in range(n - 1):
if g[i][j] == '1':
if g[i + 1][j] != '1' and g[i][j + 1] != '1':
flag = 1
break
if flag:
break
if flag:
print("NO")
else:
print("YES")
```
| 94,559 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Polygon is not only the best platform for developing problems but also a square matrix with side n, initially filled with the character 0.
On the polygon, military training was held. The soldiers placed a cannon above each cell in the first row and a cannon to the left of each cell in the first column. Thus, exactly 2n cannons were placed.
<image> Initial polygon for n=4.
Cannons shoot character 1. At any moment of time, no more than one cannon is shooting. When a 1 flies out of a cannon, it flies forward (in the direction of the shot) until it collides with a polygon border or another 1. After that, it takes the cell in which it was before the collision and remains there. Take a look at the examples for better understanding.
More formally:
* if a cannon stands in the row i, to the left of the first column, and shoots with a 1, then the 1 starts its flight from the cell (i, 1) and ends in some cell (i, j);
* if a cannon stands in the column j, above the first row, and shoots with a 1, then the 1 starts its flight from the cell (1, j) and ends in some cell (i, j).
For example, consider the following sequence of shots:
<image>
1. Shoot the cannon in the row 2. 2. Shoot the cannon in the row 2. 3. Shoot the cannon in column 3.
You have a report from the military training on your desk. This report is a square matrix with side length n consisting of 0 and 1. You wonder if the training actually happened. In other words, is there a sequence of shots such that, after the training, you get the given matrix?
Each cannon can make an arbitrary number of shots. Before the training, each cell of the polygon contains 0.
Input
The first line contains an integer t (1 β€ t β€ 1000) β the number of test cases. Then t test cases follow.
Each test case starts with a line containing an integer n (1 β€ n β€ 50) β the size of the polygon.
This is followed by n lines of length n, consisting of 0 and 1 β the polygon matrix after the training.
The total area of the matrices in all test cases in one test does not exceed 10^5.
Output
For each test case print:
* YES if there is a sequence of shots leading to a given matrix;
* NO if such a sequence does not exist.
The letters in the words YES and NO can be printed in any case.
Example
Input
5
4
0010
0011
0000
0000
2
10
01
2
00
00
4
0101
1111
0101
0111
4
0100
1110
0101
0111
Output
YES
NO
YES
YES
NO
Note
The first test case was explained in the statement.
The answer to the second test case is NO, since a 1 in a cell (1, 1) flying out of any cannon would continue its flight further.
Tags: dp, graphs, implementation, shortest paths
Correct Solution:
```
#https://codeforces.com/contest/1360/problem/E
n_data = int(input())
datas = []
for i in range(n_data):
l_array = int(input())
tab = []
for i in range(l_array):
row = input()
r = []
for c in row: r.append(c)
tab.append(r)
datas.append(tab)
#for d in datas:
# print(d)
def is_border(x, y, grid):
if x == len(grid[0])-1: return True
if y == len(grid)-1: return True
return False
def possible(x, y, grid):
if is_border(x, y, grid): return True
if grid[x+1][y] == '1': return True
if grid[x][y+1] == '1': return True
return False
def grid_OK(grid):
for y in range(len(grid)):
for x in range(len(grid[0])):
if grid[x][y] == '0': continue
if possible(x, y, grid): continue
return 'NO'
return 'YES'
for d in datas:
print(grid_OK(d))
```
| 94,560 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polygon is not only the best platform for developing problems but also a square matrix with side n, initially filled with the character 0.
On the polygon, military training was held. The soldiers placed a cannon above each cell in the first row and a cannon to the left of each cell in the first column. Thus, exactly 2n cannons were placed.
<image> Initial polygon for n=4.
Cannons shoot character 1. At any moment of time, no more than one cannon is shooting. When a 1 flies out of a cannon, it flies forward (in the direction of the shot) until it collides with a polygon border or another 1. After that, it takes the cell in which it was before the collision and remains there. Take a look at the examples for better understanding.
More formally:
* if a cannon stands in the row i, to the left of the first column, and shoots with a 1, then the 1 starts its flight from the cell (i, 1) and ends in some cell (i, j);
* if a cannon stands in the column j, above the first row, and shoots with a 1, then the 1 starts its flight from the cell (1, j) and ends in some cell (i, j).
For example, consider the following sequence of shots:
<image>
1. Shoot the cannon in the row 2. 2. Shoot the cannon in the row 2. 3. Shoot the cannon in column 3.
You have a report from the military training on your desk. This report is a square matrix with side length n consisting of 0 and 1. You wonder if the training actually happened. In other words, is there a sequence of shots such that, after the training, you get the given matrix?
Each cannon can make an arbitrary number of shots. Before the training, each cell of the polygon contains 0.
Input
The first line contains an integer t (1 β€ t β€ 1000) β the number of test cases. Then t test cases follow.
Each test case starts with a line containing an integer n (1 β€ n β€ 50) β the size of the polygon.
This is followed by n lines of length n, consisting of 0 and 1 β the polygon matrix after the training.
The total area of the matrices in all test cases in one test does not exceed 10^5.
Output
For each test case print:
* YES if there is a sequence of shots leading to a given matrix;
* NO if such a sequence does not exist.
The letters in the words YES and NO can be printed in any case.
Example
Input
5
4
0010
0011
0000
0000
2
10
01
2
00
00
4
0101
1111
0101
0111
4
0100
1110
0101
0111
Output
YES
NO
YES
YES
NO
Note
The first test case was explained in the statement.
The answer to the second test case is NO, since a 1 in a cell (1, 1) flying out of any cannon would continue its flight further.
Submitted Solution:
```
class Solution():
def __init__(self):
test = int(input())
for i in range(0, test):
n = int(input())
a = [input() for i in range(0, n)]
print(self.solve(n, a))
def solve(self, n, a):
q = [None for i in range(0, 60 * 60)]
used = [[False for i in range(0, n)] for j in range(0, n)]
l = r = 0
p = [(0, -1), (-1, 0)]
for i in range(0, n):
if a[i][n - 1] == '1':
q[r] = (i, n - 1)
r += 1
used[i][n - 1] = True
if a[n - 1][i] == '1' and i < n - 1:
q[r] = (n - 1, i)
r += 1
used[n - 1][i] = True
while l < r:
u = q[l]
l += 1
for i in p:
v = (u[0] + i[0], u[1] + i[1])
if min(v[0], v[1]) >= 0 and max(v[0], v[1]) < n and not used[v[0]][v[1]] and a[v[0]][v[1]] == '1':
q[r] = v
r += 1
used[v[0]][v[1]] = True
for i in range(0, n):
for j in range(0, n):
if a[i][j] == '1' and not used[i][j]:
return "NO"
return "YES"
Solution()
```
Yes
| 94,561 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polygon is not only the best platform for developing problems but also a square matrix with side n, initially filled with the character 0.
On the polygon, military training was held. The soldiers placed a cannon above each cell in the first row and a cannon to the left of each cell in the first column. Thus, exactly 2n cannons were placed.
<image> Initial polygon for n=4.
Cannons shoot character 1. At any moment of time, no more than one cannon is shooting. When a 1 flies out of a cannon, it flies forward (in the direction of the shot) until it collides with a polygon border or another 1. After that, it takes the cell in which it was before the collision and remains there. Take a look at the examples for better understanding.
More formally:
* if a cannon stands in the row i, to the left of the first column, and shoots with a 1, then the 1 starts its flight from the cell (i, 1) and ends in some cell (i, j);
* if a cannon stands in the column j, above the first row, and shoots with a 1, then the 1 starts its flight from the cell (1, j) and ends in some cell (i, j).
For example, consider the following sequence of shots:
<image>
1. Shoot the cannon in the row 2. 2. Shoot the cannon in the row 2. 3. Shoot the cannon in column 3.
You have a report from the military training on your desk. This report is a square matrix with side length n consisting of 0 and 1. You wonder if the training actually happened. In other words, is there a sequence of shots such that, after the training, you get the given matrix?
Each cannon can make an arbitrary number of shots. Before the training, each cell of the polygon contains 0.
Input
The first line contains an integer t (1 β€ t β€ 1000) β the number of test cases. Then t test cases follow.
Each test case starts with a line containing an integer n (1 β€ n β€ 50) β the size of the polygon.
This is followed by n lines of length n, consisting of 0 and 1 β the polygon matrix after the training.
The total area of the matrices in all test cases in one test does not exceed 10^5.
Output
For each test case print:
* YES if there is a sequence of shots leading to a given matrix;
* NO if such a sequence does not exist.
The letters in the words YES and NO can be printed in any case.
Example
Input
5
4
0010
0011
0000
0000
2
10
01
2
00
00
4
0101
1111
0101
0111
4
0100
1110
0101
0111
Output
YES
NO
YES
YES
NO
Note
The first test case was explained in the statement.
The answer to the second test case is NO, since a 1 in a cell (1, 1) flying out of any cannon would continue its flight further.
Submitted Solution:
```
T = int(input())
def fn_check(arr, n):
for i in range(n):
for j in range(n):
if (arr[i][j] == '1'):
if (i!=n-1 and j!=n-1):
if (arr[i+1][j] =='0' and arr[i][j+1]=='0'):
return "NO"
return "YES"
for t in range(T):
n = int(input())
arr = []
for i in range(n):
l = list(input())
arr.append(l)
print(fn_check(arr, n))
```
Yes
| 94,562 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polygon is not only the best platform for developing problems but also a square matrix with side n, initially filled with the character 0.
On the polygon, military training was held. The soldiers placed a cannon above each cell in the first row and a cannon to the left of each cell in the first column. Thus, exactly 2n cannons were placed.
<image> Initial polygon for n=4.
Cannons shoot character 1. At any moment of time, no more than one cannon is shooting. When a 1 flies out of a cannon, it flies forward (in the direction of the shot) until it collides with a polygon border or another 1. After that, it takes the cell in which it was before the collision and remains there. Take a look at the examples for better understanding.
More formally:
* if a cannon stands in the row i, to the left of the first column, and shoots with a 1, then the 1 starts its flight from the cell (i, 1) and ends in some cell (i, j);
* if a cannon stands in the column j, above the first row, and shoots with a 1, then the 1 starts its flight from the cell (1, j) and ends in some cell (i, j).
For example, consider the following sequence of shots:
<image>
1. Shoot the cannon in the row 2. 2. Shoot the cannon in the row 2. 3. Shoot the cannon in column 3.
You have a report from the military training on your desk. This report is a square matrix with side length n consisting of 0 and 1. You wonder if the training actually happened. In other words, is there a sequence of shots such that, after the training, you get the given matrix?
Each cannon can make an arbitrary number of shots. Before the training, each cell of the polygon contains 0.
Input
The first line contains an integer t (1 β€ t β€ 1000) β the number of test cases. Then t test cases follow.
Each test case starts with a line containing an integer n (1 β€ n β€ 50) β the size of the polygon.
This is followed by n lines of length n, consisting of 0 and 1 β the polygon matrix after the training.
The total area of the matrices in all test cases in one test does not exceed 10^5.
Output
For each test case print:
* YES if there is a sequence of shots leading to a given matrix;
* NO if such a sequence does not exist.
The letters in the words YES and NO can be printed in any case.
Example
Input
5
4
0010
0011
0000
0000
2
10
01
2
00
00
4
0101
1111
0101
0111
4
0100
1110
0101
0111
Output
YES
NO
YES
YES
NO
Note
The first test case was explained in the statement.
The answer to the second test case is NO, since a 1 in a cell (1, 1) flying out of any cannon would continue its flight further.
Submitted Solution:
```
#!/usr/bin/env python3
# coding: utf-8
import sys
input = sys.stdin.readline
def inp():
return int(input())
def inlt():
return list(map(int, input().split()))
def insr():
s = input()
return list(s[: len(s) - 1])
def invr():
return map(int, input().split())
def solve(matrix, n):
for i in range(n-1):
for j in range(n-1):
if matrix[i][j] == '1' and matrix[i+1][j] == '0' and matrix[i][j+1] == '0':
return 'No'
return 'Yes'
def main():
t = inp()
for _ in range(t):
n = inp()
matrix = [insr() for _ in range(n)]
# print(matrix)
ans = solve(matrix, n)
print(ans)
if __name__ == "__main__":
main()
```
Yes
| 94,563 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polygon is not only the best platform for developing problems but also a square matrix with side n, initially filled with the character 0.
On the polygon, military training was held. The soldiers placed a cannon above each cell in the first row and a cannon to the left of each cell in the first column. Thus, exactly 2n cannons were placed.
<image> Initial polygon for n=4.
Cannons shoot character 1. At any moment of time, no more than one cannon is shooting. When a 1 flies out of a cannon, it flies forward (in the direction of the shot) until it collides with a polygon border or another 1. After that, it takes the cell in which it was before the collision and remains there. Take a look at the examples for better understanding.
More formally:
* if a cannon stands in the row i, to the left of the first column, and shoots with a 1, then the 1 starts its flight from the cell (i, 1) and ends in some cell (i, j);
* if a cannon stands in the column j, above the first row, and shoots with a 1, then the 1 starts its flight from the cell (1, j) and ends in some cell (i, j).
For example, consider the following sequence of shots:
<image>
1. Shoot the cannon in the row 2. 2. Shoot the cannon in the row 2. 3. Shoot the cannon in column 3.
You have a report from the military training on your desk. This report is a square matrix with side length n consisting of 0 and 1. You wonder if the training actually happened. In other words, is there a sequence of shots such that, after the training, you get the given matrix?
Each cannon can make an arbitrary number of shots. Before the training, each cell of the polygon contains 0.
Input
The first line contains an integer t (1 β€ t β€ 1000) β the number of test cases. Then t test cases follow.
Each test case starts with a line containing an integer n (1 β€ n β€ 50) β the size of the polygon.
This is followed by n lines of length n, consisting of 0 and 1 β the polygon matrix after the training.
The total area of the matrices in all test cases in one test does not exceed 10^5.
Output
For each test case print:
* YES if there is a sequence of shots leading to a given matrix;
* NO if such a sequence does not exist.
The letters in the words YES and NO can be printed in any case.
Example
Input
5
4
0010
0011
0000
0000
2
10
01
2
00
00
4
0101
1111
0101
0111
4
0100
1110
0101
0111
Output
YES
NO
YES
YES
NO
Note
The first test case was explained in the statement.
The answer to the second test case is NO, since a 1 in a cell (1, 1) flying out of any cannon would continue its flight further.
Submitted Solution:
```
def solve(A,n):
for i in range(n-1):
for j in range(n-1):
if(A[i][j]=='1'):
if(A[i][j+1]!='1' and A[i+1][j]!='1'):
print("NO")
return
print("YES")
t=int(input())
for _ in range(t):
n=int(input())
A=[]
for i in range(n):
A.append(input())
if (n == 1):
print("YES")
continue
solve(A,n)
```
Yes
| 94,564 |
Evaluate the correctness of the submitted Python 2 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polygon is not only the best platform for developing problems but also a square matrix with side n, initially filled with the character 0.
On the polygon, military training was held. The soldiers placed a cannon above each cell in the first row and a cannon to the left of each cell in the first column. Thus, exactly 2n cannons were placed.
<image> Initial polygon for n=4.
Cannons shoot character 1. At any moment of time, no more than one cannon is shooting. When a 1 flies out of a cannon, it flies forward (in the direction of the shot) until it collides with a polygon border or another 1. After that, it takes the cell in which it was before the collision and remains there. Take a look at the examples for better understanding.
More formally:
* if a cannon stands in the row i, to the left of the first column, and shoots with a 1, then the 1 starts its flight from the cell (i, 1) and ends in some cell (i, j);
* if a cannon stands in the column j, above the first row, and shoots with a 1, then the 1 starts its flight from the cell (1, j) and ends in some cell (i, j).
For example, consider the following sequence of shots:
<image>
1. Shoot the cannon in the row 2. 2. Shoot the cannon in the row 2. 3. Shoot the cannon in column 3.
You have a report from the military training on your desk. This report is a square matrix with side length n consisting of 0 and 1. You wonder if the training actually happened. In other words, is there a sequence of shots such that, after the training, you get the given matrix?
Each cannon can make an arbitrary number of shots. Before the training, each cell of the polygon contains 0.
Input
The first line contains an integer t (1 β€ t β€ 1000) β the number of test cases. Then t test cases follow.
Each test case starts with a line containing an integer n (1 β€ n β€ 50) β the size of the polygon.
This is followed by n lines of length n, consisting of 0 and 1 β the polygon matrix after the training.
The total area of the matrices in all test cases in one test does not exceed 10^5.
Output
For each test case print:
* YES if there is a sequence of shots leading to a given matrix;
* NO if such a sequence does not exist.
The letters in the words YES and NO can be printed in any case.
Example
Input
5
4
0010
0011
0000
0000
2
10
01
2
00
00
4
0101
1111
0101
0111
4
0100
1110
0101
0111
Output
YES
NO
YES
YES
NO
Note
The first test case was explained in the statement.
The answer to the second test case is NO, since a 1 in a cell (1, 1) flying out of any cannon would continue its flight further.
Submitted Solution:
```
from sys import stdin, stdout
from collections import Counter, defaultdict
from itertools import permutations, combinations
raw_input = stdin.readline
pr = stdout.write
mod=10**9+7
def ni():
return int(raw_input())
def li():
return map(int,raw_input().strip())
def pn(n):
stdout.write(str(n)+'\n')
def pa(arr):
pr(' '.join(map(str,arr))+'\n')
# fast read function for total integer input
def inp():
# this function returns whole input of
# space/line seperated integers
# Use Ctrl+D to flush stdin.
return map(int,stdin.read().split())
range = xrange # not for python 3.0+
# main code
pre=[[0 for i in range(51)] for j in range(51)]
suf=[[0 for i in range(51)] for j in range(51)]
for t in range(input()):
n=ni()
mat=[]
for i in range(n):
mat.append(list(reversed(li())))
mat=list(reversed(mat))
f=0
for i in range(n):
for j in range(n):
if mat[i][j]:
if not ( (i!=0 and mat[i-1][j]) or (j!=0 and mat[i][j-1])):
if i!=0 and j!=0:
f=1
break
if f:
break
if f:
pr('NO\n')
else:
pr('YES\n')
```
Yes
| 94,565 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polygon is not only the best platform for developing problems but also a square matrix with side n, initially filled with the character 0.
On the polygon, military training was held. The soldiers placed a cannon above each cell in the first row and a cannon to the left of each cell in the first column. Thus, exactly 2n cannons were placed.
<image> Initial polygon for n=4.
Cannons shoot character 1. At any moment of time, no more than one cannon is shooting. When a 1 flies out of a cannon, it flies forward (in the direction of the shot) until it collides with a polygon border or another 1. After that, it takes the cell in which it was before the collision and remains there. Take a look at the examples for better understanding.
More formally:
* if a cannon stands in the row i, to the left of the first column, and shoots with a 1, then the 1 starts its flight from the cell (i, 1) and ends in some cell (i, j);
* if a cannon stands in the column j, above the first row, and shoots with a 1, then the 1 starts its flight from the cell (1, j) and ends in some cell (i, j).
For example, consider the following sequence of shots:
<image>
1. Shoot the cannon in the row 2. 2. Shoot the cannon in the row 2. 3. Shoot the cannon in column 3.
You have a report from the military training on your desk. This report is a square matrix with side length n consisting of 0 and 1. You wonder if the training actually happened. In other words, is there a sequence of shots such that, after the training, you get the given matrix?
Each cannon can make an arbitrary number of shots. Before the training, each cell of the polygon contains 0.
Input
The first line contains an integer t (1 β€ t β€ 1000) β the number of test cases. Then t test cases follow.
Each test case starts with a line containing an integer n (1 β€ n β€ 50) β the size of the polygon.
This is followed by n lines of length n, consisting of 0 and 1 β the polygon matrix after the training.
The total area of the matrices in all test cases in one test does not exceed 10^5.
Output
For each test case print:
* YES if there is a sequence of shots leading to a given matrix;
* NO if such a sequence does not exist.
The letters in the words YES and NO can be printed in any case.
Example
Input
5
4
0010
0011
0000
0000
2
10
01
2
00
00
4
0101
1111
0101
0111
4
0100
1110
0101
0111
Output
YES
NO
YES
YES
NO
Note
The first test case was explained in the statement.
The answer to the second test case is NO, since a 1 in a cell (1, 1) flying out of any cannon would continue its flight further.
Submitted Solution:
```
#------------------------------what is this I don't know....just makes my mess faster--------------------------------------
import os
import sys
from io import BytesIO, IOBase
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
#----------------------------------Real game starts here--------------------------------------
'''
___________________THIS IS AESTROIX CODE________________________
KARMANYA GUPTA
'''
#_______________________________________________________________#
def fact(x):
if x == 0:
return 1
else:
return x * fact(x-1)
def lower_bound(li, num): #return 0 if all are greater or equal to
answer = -1
start = 0
end = len(li)-1
while(start <= end):
middle = (end+start)//2
if li[middle] >= num:
answer = middle
end = middle - 1
else:
start = middle + 1
return answer #index where x is not less than num
def upper_bound(li, num): #return n-1 if all are small or equal
answer = -1
start = 0
end = len(li)-1
while(start <= end):
middle = (end+start)//2
if li[middle] <= num:
answer = middle
start = middle + 1
else:
end = middle - 1
return answer #index where x is not greater than num
def abs(x):
return x if x >=0 else -x
def binary_search(li, val, lb, ub):
ans = 0
while(lb <= ub):
mid = (lb+ub)//2
#print(mid, li[mid])
if li[mid] > val:
ub = mid-1
elif val > li[mid]:
lb = mid + 1
else:
ans = 1
break
return ans
#_______________________________________________________________#
from math import *
for _ in range(int(input())):
n = int(input())
if n==1:
print("YES")
continue
mat = []
for i in range(n):
string = input()
li = []
for j in range(n):
li.append(int(string[j]))
mat.append(li)
flag = 1
for i in range(n-1):
for j in range(n-1):
if mat[i][j] == 1:
if mat[i+1][j] == 0 and mat[i][j+1] == 0:
flag = 0
break
"""for j in range(n-1):
if mat[n-1][j] == 1 and mat[n-1][j+1] == 0:
print("there.......")
flag = 0
break
for i in range(n-1):
if mat[i][n-1] == 1 and mat[i+1][n-1] == 0:
print("thence.....")
flag = 0
break"""
print("YNEOS"[flag==0::2])
```
No
| 94,566 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polygon is not only the best platform for developing problems but also a square matrix with side n, initially filled with the character 0.
On the polygon, military training was held. The soldiers placed a cannon above each cell in the first row and a cannon to the left of each cell in the first column. Thus, exactly 2n cannons were placed.
<image> Initial polygon for n=4.
Cannons shoot character 1. At any moment of time, no more than one cannon is shooting. When a 1 flies out of a cannon, it flies forward (in the direction of the shot) until it collides with a polygon border or another 1. After that, it takes the cell in which it was before the collision and remains there. Take a look at the examples for better understanding.
More formally:
* if a cannon stands in the row i, to the left of the first column, and shoots with a 1, then the 1 starts its flight from the cell (i, 1) and ends in some cell (i, j);
* if a cannon stands in the column j, above the first row, and shoots with a 1, then the 1 starts its flight from the cell (1, j) and ends in some cell (i, j).
For example, consider the following sequence of shots:
<image>
1. Shoot the cannon in the row 2. 2. Shoot the cannon in the row 2. 3. Shoot the cannon in column 3.
You have a report from the military training on your desk. This report is a square matrix with side length n consisting of 0 and 1. You wonder if the training actually happened. In other words, is there a sequence of shots such that, after the training, you get the given matrix?
Each cannon can make an arbitrary number of shots. Before the training, each cell of the polygon contains 0.
Input
The first line contains an integer t (1 β€ t β€ 1000) β the number of test cases. Then t test cases follow.
Each test case starts with a line containing an integer n (1 β€ n β€ 50) β the size of the polygon.
This is followed by n lines of length n, consisting of 0 and 1 β the polygon matrix after the training.
The total area of the matrices in all test cases in one test does not exceed 10^5.
Output
For each test case print:
* YES if there is a sequence of shots leading to a given matrix;
* NO if such a sequence does not exist.
The letters in the words YES and NO can be printed in any case.
Example
Input
5
4
0010
0011
0000
0000
2
10
01
2
00
00
4
0101
1111
0101
0111
4
0100
1110
0101
0111
Output
YES
NO
YES
YES
NO
Note
The first test case was explained in the statement.
The answer to the second test case is NO, since a 1 in a cell (1, 1) flying out of any cannon would continue its flight further.
Submitted Solution:
```
for _ in range(int(input())):
n = int(input())
l = []
for i in range(n):
l.append(input())
# print(l)
flag = 0
for i in range(n):
for j in range(n):
if i==n-1:
if l[i][j]=='1' and j<n-1:
if l[i][j+1] == '0':
flag = 1
break
else:
if l[i][j]=='1' and j<n-1:
if l[i+1][j] == '0' and l[i][j+1] == '0':
flag = 1
break
if flag == 1:
break
if flag==1:
print('NO')
else:
print('YES')
```
No
| 94,567 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polygon is not only the best platform for developing problems but also a square matrix with side n, initially filled with the character 0.
On the polygon, military training was held. The soldiers placed a cannon above each cell in the first row and a cannon to the left of each cell in the first column. Thus, exactly 2n cannons were placed.
<image> Initial polygon for n=4.
Cannons shoot character 1. At any moment of time, no more than one cannon is shooting. When a 1 flies out of a cannon, it flies forward (in the direction of the shot) until it collides with a polygon border or another 1. After that, it takes the cell in which it was before the collision and remains there. Take a look at the examples for better understanding.
More formally:
* if a cannon stands in the row i, to the left of the first column, and shoots with a 1, then the 1 starts its flight from the cell (i, 1) and ends in some cell (i, j);
* if a cannon stands in the column j, above the first row, and shoots with a 1, then the 1 starts its flight from the cell (1, j) and ends in some cell (i, j).
For example, consider the following sequence of shots:
<image>
1. Shoot the cannon in the row 2. 2. Shoot the cannon in the row 2. 3. Shoot the cannon in column 3.
You have a report from the military training on your desk. This report is a square matrix with side length n consisting of 0 and 1. You wonder if the training actually happened. In other words, is there a sequence of shots such that, after the training, you get the given matrix?
Each cannon can make an arbitrary number of shots. Before the training, each cell of the polygon contains 0.
Input
The first line contains an integer t (1 β€ t β€ 1000) β the number of test cases. Then t test cases follow.
Each test case starts with a line containing an integer n (1 β€ n β€ 50) β the size of the polygon.
This is followed by n lines of length n, consisting of 0 and 1 β the polygon matrix after the training.
The total area of the matrices in all test cases in one test does not exceed 10^5.
Output
For each test case print:
* YES if there is a sequence of shots leading to a given matrix;
* NO if such a sequence does not exist.
The letters in the words YES and NO can be printed in any case.
Example
Input
5
4
0010
0011
0000
0000
2
10
01
2
00
00
4
0101
1111
0101
0111
4
0100
1110
0101
0111
Output
YES
NO
YES
YES
NO
Note
The first test case was explained in the statement.
The answer to the second test case is NO, since a 1 in a cell (1, 1) flying out of any cannon would continue its flight further.
Submitted Solution:
```
import sys
import string
from collections import Counter, defaultdict
from math import fsum, sqrt, gcd, ceil, factorial
from itertools import combinations,permutations
# input = sys.stdin.readline
flush = lambda : sys.stdout.flush
comb = lambda x , y : (factorial(x) // factorial(y)) // factorial(x - y)
#inputs
# ip = lambda : input().rstrip()
ip = lambda : input()
ii = lambda : int(input())
r = lambda : map(int, input().split())
rr = lambda : list(r())
```
No
| 94,568 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polygon is not only the best platform for developing problems but also a square matrix with side n, initially filled with the character 0.
On the polygon, military training was held. The soldiers placed a cannon above each cell in the first row and a cannon to the left of each cell in the first column. Thus, exactly 2n cannons were placed.
<image> Initial polygon for n=4.
Cannons shoot character 1. At any moment of time, no more than one cannon is shooting. When a 1 flies out of a cannon, it flies forward (in the direction of the shot) until it collides with a polygon border or another 1. After that, it takes the cell in which it was before the collision and remains there. Take a look at the examples for better understanding.
More formally:
* if a cannon stands in the row i, to the left of the first column, and shoots with a 1, then the 1 starts its flight from the cell (i, 1) and ends in some cell (i, j);
* if a cannon stands in the column j, above the first row, and shoots with a 1, then the 1 starts its flight from the cell (1, j) and ends in some cell (i, j).
For example, consider the following sequence of shots:
<image>
1. Shoot the cannon in the row 2. 2. Shoot the cannon in the row 2. 3. Shoot the cannon in column 3.
You have a report from the military training on your desk. This report is a square matrix with side length n consisting of 0 and 1. You wonder if the training actually happened. In other words, is there a sequence of shots such that, after the training, you get the given matrix?
Each cannon can make an arbitrary number of shots. Before the training, each cell of the polygon contains 0.
Input
The first line contains an integer t (1 β€ t β€ 1000) β the number of test cases. Then t test cases follow.
Each test case starts with a line containing an integer n (1 β€ n β€ 50) β the size of the polygon.
This is followed by n lines of length n, consisting of 0 and 1 β the polygon matrix after the training.
The total area of the matrices in all test cases in one test does not exceed 10^5.
Output
For each test case print:
* YES if there is a sequence of shots leading to a given matrix;
* NO if such a sequence does not exist.
The letters in the words YES and NO can be printed in any case.
Example
Input
5
4
0010
0011
0000
0000
2
10
01
2
00
00
4
0101
1111
0101
0111
4
0100
1110
0101
0111
Output
YES
NO
YES
YES
NO
Note
The first test case was explained in the statement.
The answer to the second test case is NO, since a 1 in a cell (1, 1) flying out of any cannon would continue its flight further.
Submitted Solution:
```
def find_training():
one = []
edge = []
inside = []
n = int(input())
for i in range(n):
row = input()
row = list(row)
row = list(map(int, row))
for idx,val in enumerate(row):
if val == 1:
one.append([i,idx])
if idx == n-1:
edge.append([i,idx])
else:
inside.append([i,idx])
result = 'YES'
for element in inside:
if [element[0]+1, element[1]] in one or [element[0], element[1]+1] in one:
result = 'YES'
else:
result = 'NO'
break
return result
t = int(input())
output = []
for i in range(t):
output.append(find_training())
for i in output:
print(i)
```
No
| 94,569 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let a and b be some non-negative integers. Let's define strange addition of a and b as following:
1. write down the numbers one under another and align them by their least significant digit;
2. add them up digit by digit and concatenate the respective sums together.
Assume that both numbers have an infinite number of leading zeros.
For example, let's take a look at a strange addition of numbers 3248 and 908:
<image>
You are given a string c, consisting of n digits from 0 to 9. You are also given m updates of form:
* x~d β replace the digit at the x-th position of c with a digit d.
Note that string c might have leading zeros at any point of time.
After each update print the number of pairs (a, b) such that both a and b are non-negative integers and the result of a strange addition of a and b is equal to c.
Note that the numbers of pairs can be quite large, so print them modulo 998244353.
Input
The first line contains two integers n and m (1 β€ n, m β€ 5 β
10^5) β the length of the number c and the number of updates.
The second line contains a string c, consisting of exactly n digits from 0 to 9.
Each of the next m lines contains two integers x and d (1 β€ x β€ n, 0 β€ d β€ 9) β the descriptions of updates.
Output
Print m integers β the i-th value should be equal to the number of pairs (a, b) such that both a and b are non-negative integers and the result of a strange addition of a and b is equal to c after i updates are applied.
Note that the numbers of pairs can be quite large, so print them modulo 998244353.
Example
Input
2 3
14
2 4
2 1
1 0
Output
15
12
2
Note
After the first update c is equal to 14. The pairs that sum up to 14 are: (0, 14), (1, 13), (2, 12), (3, 11), (4, 10), (5, 9), (6, 8), (7, 7), (8, 6), (9, 5), (10, 4), (11, 3), (12, 2), (13, 1), (14, 0).
After the second update c is equal to 11.
After the third update c is equal to 01.
Tags: data structures, dp, matrices
Correct Solution:
```
from sys import stdin
import sys
class LazySegTree:
def __init__(self,N,first):
self.LV = (N-1).bit_length()
self.NO = 2**self.LV
self.First = first
self.data = [first] * (2*self.NO)
self.lazy = [None] * (2*self.NO)
def calc(self,l,r):
return min(l,r)
def gindex(self,l,r):
L = l + self.NO
R = r + self.NO
lm = (L // (L & -L)) >> 1
rm = (R // (R & -R)) >> 1
while L < R:
if R <= rm:
yield R
if L <= lm:
yield L
L >>= 1; R >>= 1
while L:
yield L
L >>= 1
def propagates(self,*ids):
for i in reversed(ids):
v = self.lazy[i-1]
if v is None:
continue
self.lazy[2*i-1] = self.data[2*i-1] = self.lazy[2*i] = self.data[2*i] = v
self.lazy[i-1] = None
def update(self , l, r, x):
*ids, = self.gindex(l, r)
self.propagates(*ids)
L = self.NO + l; R = self.NO + r
while L < R:
if R & 1:
R -= 1
self.lazy[R-1] = self.data[R-1] = x
if L & 1:
self.lazy[L-1] = self.data[L-1] = x
L += 1
L >>= 1; R >>= 1
for i in ids:
self.data[i-1] = self.calc(self.data[2*i-1], self.data[2*i])
def query(self , l, r):
self.propagates(*self.gindex(l, r))
L = self.NO + l; R = self.NO + r
s = self.First
while L < R:
if R & 1:
R -= 1
s = self.calc(s, self.data[R-1])
if L & 1:
s = self.calc(s, self.data[L-1])
L += 1
L >>= 1; R >>= 1
return s
def inverse(a,mod):
return pow(a,mod-2,mod)
def calc(tmpR):
ret = dp[tmpR-Lside[tmpR]] * dnum[c[tmpR]]
if tmpR - Lside[tmpR] > 0:
ret += dp[tmpR-Lside[tmpR]-1] * dnum[c[tmpR] + 10]
return ret
mod = 998244353
n,m = map(int,stdin.readline().split())
c = list(stdin.readline()[:-1])
c = list(map(int,c))
c = [0] + c + [9]
#divide
dnum = [0] * 20
for i in range(10):
for j in range(10):
dnum[i+j] += 1
#dp
dp = [0] * (n+10)
dp[0] = 1
for i in range(n+8):
dp[i+1] += dp[i] * 2
dp[i+1] %= mod
dp[i+2] += dp[i] * 8
dp[i+2] %= mod
#construct
ST = LazySegTree(len(c),float("inf"))
Lside = [None] * len(c)
tmpR = None
ans = 1
for i in range(len(c)-1,-1,-1):
if c[i] != 1:
if tmpR != None:
ST.update(i+1,tmpR+1,tmpR)
Lside[tmpR] = i+1
#now = dp[tmpR-i-1] * dnum[c[tmpR]]
#if tmpR-i-1 > 0:
# now += dp[tmpR-i-2] * dnum[c[tmpR] + 10]
ans *= calc(tmpR)
ans %= mod
tmpR = i
print (ans * inverse(dnum[9] , mod) % mod , file=sys.stderr)
anss = []
vvv = inverse(dnum[9] , mod)
for loop in range(m):
x,d = map(int,stdin.readline().split())
if d == 1:
if c[x] != 1:
now = 1
newR = ST.query(x+1,x+2)
now *= inverse(calc(x) * calc(newR),mod)
ST.update(Lside[x],x+1,newR)
c[x] = d
Lside[newR] = Lside[x]
now *= calc(newR)
ans *= now
else:
if c[x] != 1:
now = 1
now *= inverse(calc(x) , mod)
c[x] = d
now *= calc(x)
ans *= now
else:
now = 1
oldR = ST.query(x,x+1)
now *= inverse(calc(oldR) , mod)
c[x] = d
Lside[x] = Lside[oldR]
Lside[oldR] = x+1
ST.update(Lside[x],x+1,x)
now *= calc(oldR) * calc(x)
ans *= now
ans %= mod
#print (ans * inverse(dnum[9] , mod) % mod )
anss.append(ans * vvv % mod)
print ("\n".join(map(str,anss)))
```
| 94,570 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let a and b be some non-negative integers. Let's define strange addition of a and b as following:
1. write down the numbers one under another and align them by their least significant digit;
2. add them up digit by digit and concatenate the respective sums together.
Assume that both numbers have an infinite number of leading zeros.
For example, let's take a look at a strange addition of numbers 3248 and 908:
<image>
You are given a string c, consisting of n digits from 0 to 9. You are also given m updates of form:
* x~d β replace the digit at the x-th position of c with a digit d.
Note that string c might have leading zeros at any point of time.
After each update print the number of pairs (a, b) such that both a and b are non-negative integers and the result of a strange addition of a and b is equal to c.
Note that the numbers of pairs can be quite large, so print them modulo 998244353.
Input
The first line contains two integers n and m (1 β€ n, m β€ 5 β
10^5) β the length of the number c and the number of updates.
The second line contains a string c, consisting of exactly n digits from 0 to 9.
Each of the next m lines contains two integers x and d (1 β€ x β€ n, 0 β€ d β€ 9) β the descriptions of updates.
Output
Print m integers β the i-th value should be equal to the number of pairs (a, b) such that both a and b are non-negative integers and the result of a strange addition of a and b is equal to c after i updates are applied.
Note that the numbers of pairs can be quite large, so print them modulo 998244353.
Example
Input
2 3
14
2 4
2 1
1 0
Output
15
12
2
Note
After the first update c is equal to 14. The pairs that sum up to 14 are: (0, 14), (1, 13), (2, 12), (3, 11), (4, 10), (5, 9), (6, 8), (7, 7), (8, 6), (9, 5), (10, 4), (11, 3), (12, 2), (13, 1), (14, 0).
After the second update c is equal to 11.
After the third update c is equal to 01.
Submitted Solution:
```
no_d,test=map(int,input().split())
n=input()
n=list(n)
for t in range(test):
p,d=map(int,input().split())
n[p-1]=d
number=0
for i in n:
number=number*10+int(i)
else:
print((number%998244353)+1)
```
No
| 94,571 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let a and b be some non-negative integers. Let's define strange addition of a and b as following:
1. write down the numbers one under another and align them by their least significant digit;
2. add them up digit by digit and concatenate the respective sums together.
Assume that both numbers have an infinite number of leading zeros.
For example, let's take a look at a strange addition of numbers 3248 and 908:
<image>
You are given a string c, consisting of n digits from 0 to 9. You are also given m updates of form:
* x~d β replace the digit at the x-th position of c with a digit d.
Note that string c might have leading zeros at any point of time.
After each update print the number of pairs (a, b) such that both a and b are non-negative integers and the result of a strange addition of a and b is equal to c.
Note that the numbers of pairs can be quite large, so print them modulo 998244353.
Input
The first line contains two integers n and m (1 β€ n, m β€ 5 β
10^5) β the length of the number c and the number of updates.
The second line contains a string c, consisting of exactly n digits from 0 to 9.
Each of the next m lines contains two integers x and d (1 β€ x β€ n, 0 β€ d β€ 9) β the descriptions of updates.
Output
Print m integers β the i-th value should be equal to the number of pairs (a, b) such that both a and b are non-negative integers and the result of a strange addition of a and b is equal to c after i updates are applied.
Note that the numbers of pairs can be quite large, so print them modulo 998244353.
Example
Input
2 3
14
2 4
2 1
1 0
Output
15
12
2
Note
After the first update c is equal to 14. The pairs that sum up to 14 are: (0, 14), (1, 13), (2, 12), (3, 11), (4, 10), (5, 9), (6, 8), (7, 7), (8, 6), (9, 5), (10, 4), (11, 3), (12, 2), (13, 1), (14, 0).
After the second update c is equal to 11.
After the third update c is equal to 01.
Submitted Solution:
```
from bisect import bisect_left as bl
from bisect import bisect_right as br
from heapq import heappush,heappop,heapify
import math
from collections import *
from functools import reduce,cmp_to_key
import sys
input = sys.stdin.readline
M = mod = 998244353
def factors(n):return sorted(set(reduce(list.__add__, ([i, n//i] for i in range(1, int(n**0.5) + 1) if n % i == 0))))
def inv_mod(n):return pow(n, mod - 2, mod)
def li():return [int(i) for i in input().rstrip('\n').split()]
def st():return input().rstrip('\n')
def val():return int(input().rstrip('\n'))
def li2():return [i for i in input().rstrip('\n')]
def li3():return [int(i) for i in input().rstrip('\n')]
n, m = li()
c = li3()
i = mi = n - 1
ans = 0
for i in range(n):
ans = (ans + c[i] * pow(10,n - i - 1,mod))%mod
for i in range(m):
x, d = li()
x -= 1
temp = c[x]
c[x] = d
ans = (mod + ans - temp * pow(10,n - 1 - x,mod))%mod
ans = (ans + mod + d * pow(10,n - 1 - x,mod))%mod
print((ans + 1)%mod)
```
No
| 94,572 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let a and b be some non-negative integers. Let's define strange addition of a and b as following:
1. write down the numbers one under another and align them by their least significant digit;
2. add them up digit by digit and concatenate the respective sums together.
Assume that both numbers have an infinite number of leading zeros.
For example, let's take a look at a strange addition of numbers 3248 and 908:
<image>
You are given a string c, consisting of n digits from 0 to 9. You are also given m updates of form:
* x~d β replace the digit at the x-th position of c with a digit d.
Note that string c might have leading zeros at any point of time.
After each update print the number of pairs (a, b) such that both a and b are non-negative integers and the result of a strange addition of a and b is equal to c.
Note that the numbers of pairs can be quite large, so print them modulo 998244353.
Input
The first line contains two integers n and m (1 β€ n, m β€ 5 β
10^5) β the length of the number c and the number of updates.
The second line contains a string c, consisting of exactly n digits from 0 to 9.
Each of the next m lines contains two integers x and d (1 β€ x β€ n, 0 β€ d β€ 9) β the descriptions of updates.
Output
Print m integers β the i-th value should be equal to the number of pairs (a, b) such that both a and b are non-negative integers and the result of a strange addition of a and b is equal to c after i updates are applied.
Note that the numbers of pairs can be quite large, so print them modulo 998244353.
Example
Input
2 3
14
2 4
2 1
1 0
Output
15
12
2
Note
After the first update c is equal to 14. The pairs that sum up to 14 are: (0, 14), (1, 13), (2, 12), (3, 11), (4, 10), (5, 9), (6, 8), (7, 7), (8, 6), (9, 5), (10, 4), (11, 3), (12, 2), (13, 1), (14, 0).
After the second update c is equal to 11.
After the third update c is equal to 01.
Submitted Solution:
```
def comb(t2, t1):
a, b, c, d = t1
aa, bb, cc, dd = t2
return (a * aa + b * cc, a * bb + b * dd, c*aa + d * cc, c * bb + d * dd)
class SegmentTree:
def __init__(self, data, default=(1,0,0,1), func=comb):
"""initialize the segment tree with data"""
self._default = default
self._func = func
self._len = len(data)
self._size = _size = 1 << (self._len - 1).bit_length()
self.data = [default] * (2 * _size)
self.data[_size:_size + self._len] = data
for i in reversed(range(_size)):
self.data[i] = func(self.data[i + i], self.data[i + i + 1])
def __delitem__(self, idx):
self[idx] = self._default
def __getitem__(self, idx):
return self.data[idx + self._size]
def __setitem__(self, idx, value):
idx += self._size
self.data[idx] = value
idx >>= 1
while idx:
self.data[idx] = self._func(self.data[2 * idx], self.data[2 * idx + 1])
idx >>= 1
def __len__(self):
return self._len
def query(self, start, stop):
"""func of data[start, stop)"""
start += self._size
stop += self._size
res_left = res_right = self._default
while start < stop:
if start & 1:
res_left = self._func(res_left, self.data[start])
start += 1
if stop & 1:
stop -= 1
res_right = self._func(self.data[stop], res_right)
start >>= 1
stop >>= 1
return self._func(res_left, res_right)
def __repr__(self):
return "SegmentTree({0})".format(self.data)
import sys
input = sys.stdin.readline
base = [0] * 10
base[0] = (1, 9, 0, 0)
base[1] = (2, 8, 1, 0)
base[2] = (3, 7, 0, 0)
base[3] = (4, 6, 0, 0)
base[4] = (5, 5, 0, 0)
base[5] = (6, 4, 0, 0)
base[6] = (7, 3, 0, 0)
base[7] = (8, 2, 0, 0)
base[8] = (9, 1, 0, 0)
base[9] = (10,0, 0, 0)
n, m = map(int, input().split())
c = list(map(int, list(input().strip())))
based = [base[v] for v in c]
seg = SegmentTree(based)
out = [-1] * m
for q in range(m):
x, d = map(int, input().split())
seg[x-1] = base[d]
out[q] = seg.query(0, n)[0]
print('\n'.join(map(str,out)))
```
No
| 94,573 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let a and b be some non-negative integers. Let's define strange addition of a and b as following:
1. write down the numbers one under another and align them by their least significant digit;
2. add them up digit by digit and concatenate the respective sums together.
Assume that both numbers have an infinite number of leading zeros.
For example, let's take a look at a strange addition of numbers 3248 and 908:
<image>
You are given a string c, consisting of n digits from 0 to 9. You are also given m updates of form:
* x~d β replace the digit at the x-th position of c with a digit d.
Note that string c might have leading zeros at any point of time.
After each update print the number of pairs (a, b) such that both a and b are non-negative integers and the result of a strange addition of a and b is equal to c.
Note that the numbers of pairs can be quite large, so print them modulo 998244353.
Input
The first line contains two integers n and m (1 β€ n, m β€ 5 β
10^5) β the length of the number c and the number of updates.
The second line contains a string c, consisting of exactly n digits from 0 to 9.
Each of the next m lines contains two integers x and d (1 β€ x β€ n, 0 β€ d β€ 9) β the descriptions of updates.
Output
Print m integers β the i-th value should be equal to the number of pairs (a, b) such that both a and b are non-negative integers and the result of a strange addition of a and b is equal to c after i updates are applied.
Note that the numbers of pairs can be quite large, so print them modulo 998244353.
Example
Input
2 3
14
2 4
2 1
1 0
Output
15
12
2
Note
After the first update c is equal to 14. The pairs that sum up to 14 are: (0, 14), (1, 13), (2, 12), (3, 11), (4, 10), (5, 9), (6, 8), (7, 7), (8, 6), (9, 5), (10, 4), (11, 3), (12, 2), (13, 1), (14, 0).
After the second update c is equal to 11.
After the third update c is equal to 01.
Submitted Solution:
```
n, m = map(int, input().split())
c = input()
for i in range(m):
x, d = map(int, input().split())
c = c[:x-1] + str(d) + c[x:]
result = int(c)
print((result + 1)%998244353)
```
No
| 94,574 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an integer value x and a string s consisting of digits from 1 to 9 inclusive.
A substring of a string is a contiguous subsequence of that string.
Let f(l, r) be the sum of digits of a substring s[l..r].
Let's call substring s[l_1..r_1] x-prime if
* f(l_1, r_1) = x;
* there are no values l_2, r_2 such that
* l_1 β€ l_2 β€ r_2 β€ r_1;
* f(l_2, r_2) β x;
* x is divisible by f(l_2, r_2).
You are allowed to erase some characters from the string. If you erase a character, the two resulting parts of the string are concatenated without changing their order.
What is the minimum number of characters you should erase from the string so that there are no x-prime substrings in it? If there are no x-prime substrings in the given string s, then print 0.
Input
The first line contains a string s (1 β€ |s| β€ 1000). s contains only digits from 1 to 9 inclusive.
The second line contains an integer x (1 β€ x β€ 20).
Output
Print a single integer β the minimum number of characters you should erase from the string so that there are no x-prime substrings in it. If there are no x-prime substrings in the given string s, then print 0.
Examples
Input
116285317
8
Output
2
Input
314159265359
1
Output
2
Input
13
13
Output
0
Input
3434343434
7
Output
5
Note
In the first example there are two 8-prime substrings "8" and "53". You can erase these characters to get rid of both: "116285317". The resulting string "1162317" contains no 8-prime substrings. Removing these characters is also a valid answer: "116285317".
In the second example you just have to erase both ones.
In the third example there are no 13-prime substrings. There are no substrings with the sum of digits equal to 13 at all.
In the fourth example you can have neither "34", nor "43" in a string. Thus, you have to erase either all threes or all fours. There are 5 of each of them, so it doesn't matter which.
Submitted Solution:
```
s=input()
x=input()
c=0
if s.count(x)>0 :
c+=s.count(x)
print(c)
```
No
| 94,575 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an integer value x and a string s consisting of digits from 1 to 9 inclusive.
A substring of a string is a contiguous subsequence of that string.
Let f(l, r) be the sum of digits of a substring s[l..r].
Let's call substring s[l_1..r_1] x-prime if
* f(l_1, r_1) = x;
* there are no values l_2, r_2 such that
* l_1 β€ l_2 β€ r_2 β€ r_1;
* f(l_2, r_2) β x;
* x is divisible by f(l_2, r_2).
You are allowed to erase some characters from the string. If you erase a character, the two resulting parts of the string are concatenated without changing their order.
What is the minimum number of characters you should erase from the string so that there are no x-prime substrings in it? If there are no x-prime substrings in the given string s, then print 0.
Input
The first line contains a string s (1 β€ |s| β€ 1000). s contains only digits from 1 to 9 inclusive.
The second line contains an integer x (1 β€ x β€ 20).
Output
Print a single integer β the minimum number of characters you should erase from the string so that there are no x-prime substrings in it. If there are no x-prime substrings in the given string s, then print 0.
Examples
Input
116285317
8
Output
2
Input
314159265359
1
Output
2
Input
13
13
Output
0
Input
3434343434
7
Output
5
Note
In the first example there are two 8-prime substrings "8" and "53". You can erase these characters to get rid of both: "116285317". The resulting string "1162317" contains no 8-prime substrings. Removing these characters is also a valid answer: "116285317".
In the second example you just have to erase both ones.
In the third example there are no 13-prime substrings. There are no substrings with the sum of digits equal to 13 at all.
In the fourth example you can have neither "34", nor "43" in a string. Thus, you have to erase either all threes or all fours. There are 5 of each of them, so it doesn't matter which.
Submitted Solution:
```
s=input()
l=list(s)
x=eval(input())
z=0
if str(x)==s:
print(0)
else:
if str(x) in s:
z=l.count(str(x))
else:
for i in range(len(l)-1,1):
for j in range(i+1,i+2):
if (int(l[i])+int(l[j]))==x:
z=z+1
print(z)
```
No
| 94,576 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an integer value x and a string s consisting of digits from 1 to 9 inclusive.
A substring of a string is a contiguous subsequence of that string.
Let f(l, r) be the sum of digits of a substring s[l..r].
Let's call substring s[l_1..r_1] x-prime if
* f(l_1, r_1) = x;
* there are no values l_2, r_2 such that
* l_1 β€ l_2 β€ r_2 β€ r_1;
* f(l_2, r_2) β x;
* x is divisible by f(l_2, r_2).
You are allowed to erase some characters from the string. If you erase a character, the two resulting parts of the string are concatenated without changing their order.
What is the minimum number of characters you should erase from the string so that there are no x-prime substrings in it? If there are no x-prime substrings in the given string s, then print 0.
Input
The first line contains a string s (1 β€ |s| β€ 1000). s contains only digits from 1 to 9 inclusive.
The second line contains an integer x (1 β€ x β€ 20).
Output
Print a single integer β the minimum number of characters you should erase from the string so that there are no x-prime substrings in it. If there are no x-prime substrings in the given string s, then print 0.
Examples
Input
116285317
8
Output
2
Input
314159265359
1
Output
2
Input
13
13
Output
0
Input
3434343434
7
Output
5
Note
In the first example there are two 8-prime substrings "8" and "53". You can erase these characters to get rid of both: "116285317". The resulting string "1162317" contains no 8-prime substrings. Removing these characters is also a valid answer: "116285317".
In the second example you just have to erase both ones.
In the third example there are no 13-prime substrings. There are no substrings with the sum of digits equal to 13 at all.
In the fourth example you can have neither "34", nor "43" in a string. Thus, you have to erase either all threes or all fours. There are 5 of each of them, so it doesn't matter which.
Submitted Solution:
```
s=input()
l=list(s)
x=eval(input())
z=0
if str(x)==s:
print(0)
else:
if str(x) in s:
z=l.count(str(x))
else:
for i in range(len(l)-1,l):
for j in range(i+1,i+2):
if (int(l[i])+int(l[j]))==x:
z=z+1
print(z)
```
No
| 94,577 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A famous gang of pirates, Sea Dogs, has come back to their hideout from one of their extravagant plunders. They want to split their treasure fairly amongst themselves, that is why You, their trusted financial advisor, devised a game to help them:
All of them take a sit at their round table, some of them with the golden coins they have just stolen. At each iteration of the game if one of them has equal or more than 2 coins, he is eligible to the splitting and he gives one coin to each pirate sitting next to him. If there are more candidates (pirates with equal or more than 2 coins) then You are the one that chooses which one of them will do the splitting in that iteration. The game ends when there are no more candidates eligible to do the splitting.
Pirates can call it a day, only when the game ends. Since they are beings with a finite amount of time at their disposal, they would prefer if the game that they are playing can end after finite iterations, and if so, they call it a good game. On the other hand, if no matter how You do the splitting, the game cannot end in finite iterations, they call it a bad game. Can You help them figure out before they start playing if the game will be good or bad?
Input
The first line of input contains two integer numbers n and k (1 β€ n β€ 10^{9}, 0 β€ k β€ 2β
10^5), where n denotes total number of pirates and k is the number of pirates that have any coins.
The next k lines of input contain integers a_i and b_i (1 β€ a_i β€ n, 1 β€ b_i β€ 10^{9}), where a_i denotes the index of the pirate sitting at the round table (n and 1 are neighbours) and b_i the total number of coins that pirate a_i has at the start of the game.
Output
Print 1 if the game is a good game: There is a way to do the splitting so the game ends after finite number of iterations.
Print -1 if the game is a bad game: No matter how You do the splitting the game does not end in finite number of iterations.
Examples
Input
4 2
1 2
2 2
Output
1
Input
6 2
2 3
4 1
Output
1
Input
3 2
1 1
2 2
Output
-1
Note
In the third example the game has no end, because You always only have only one candidate, after whose splitting you end up in the same position as the starting one.
Tags: math
Correct Solution:
```
import os
import sys
from io import BytesIO, IOBase
def main():
pass
# region fastio
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
n, k = map(int, input().split())
coins = 0
pos = 0
for _ in range(k):
a, b = map(int, input().split())
coins += b
pos += a * b
pos %= n
if coins < n or coins == n and (pos - (n*n-n)//2) % n == 0:
print(1)
else:
print(-1)
```
| 94,578 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A famous gang of pirates, Sea Dogs, has come back to their hideout from one of their extravagant plunders. They want to split their treasure fairly amongst themselves, that is why You, their trusted financial advisor, devised a game to help them:
All of them take a sit at their round table, some of them with the golden coins they have just stolen. At each iteration of the game if one of them has equal or more than 2 coins, he is eligible to the splitting and he gives one coin to each pirate sitting next to him. If there are more candidates (pirates with equal or more than 2 coins) then You are the one that chooses which one of them will do the splitting in that iteration. The game ends when there are no more candidates eligible to do the splitting.
Pirates can call it a day, only when the game ends. Since they are beings with a finite amount of time at their disposal, they would prefer if the game that they are playing can end after finite iterations, and if so, they call it a good game. On the other hand, if no matter how You do the splitting, the game cannot end in finite iterations, they call it a bad game. Can You help them figure out before they start playing if the game will be good or bad?
Input
The first line of input contains two integer numbers n and k (1 β€ n β€ 10^{9}, 0 β€ k β€ 2β
10^5), where n denotes total number of pirates and k is the number of pirates that have any coins.
The next k lines of input contain integers a_i and b_i (1 β€ a_i β€ n, 1 β€ b_i β€ 10^{9}), where a_i denotes the index of the pirate sitting at the round table (n and 1 are neighbours) and b_i the total number of coins that pirate a_i has at the start of the game.
Output
Print 1 if the game is a good game: There is a way to do the splitting so the game ends after finite number of iterations.
Print -1 if the game is a bad game: No matter how You do the splitting the game does not end in finite number of iterations.
Examples
Input
4 2
1 2
2 2
Output
1
Input
6 2
2 3
4 1
Output
1
Input
3 2
1 1
2 2
Output
-1
Note
In the third example the game has no end, because You always only have only one candidate, after whose splitting you end up in the same position as the starting one.
Tags: math
Correct Solution:
```
import os
import sys
from io import BytesIO, IOBase
# region fastio
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
def main():
n, k = [int(x) for x in input().split()]
p, s = 0, 0
for _ in range(k):
a, b = [int(x) for x in input().split()]
s += b
p += a * b
p %= n
print(['-1', '1'][s < n or (s == n and p == (n * (n + 1) // 2) % n)])
main()
```
| 94,579 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A famous gang of pirates, Sea Dogs, has come back to their hideout from one of their extravagant plunders. They want to split their treasure fairly amongst themselves, that is why You, their trusted financial advisor, devised a game to help them:
All of them take a sit at their round table, some of them with the golden coins they have just stolen. At each iteration of the game if one of them has equal or more than 2 coins, he is eligible to the splitting and he gives one coin to each pirate sitting next to him. If there are more candidates (pirates with equal or more than 2 coins) then You are the one that chooses which one of them will do the splitting in that iteration. The game ends when there are no more candidates eligible to do the splitting.
Pirates can call it a day, only when the game ends. Since they are beings with a finite amount of time at their disposal, they would prefer if the game that they are playing can end after finite iterations, and if so, they call it a good game. On the other hand, if no matter how You do the splitting, the game cannot end in finite iterations, they call it a bad game. Can You help them figure out before they start playing if the game will be good or bad?
Input
The first line of input contains two integer numbers n and k (1 β€ n β€ 10^{9}, 0 β€ k β€ 2β
10^5), where n denotes total number of pirates and k is the number of pirates that have any coins.
The next k lines of input contain integers a_i and b_i (1 β€ a_i β€ n, 1 β€ b_i β€ 10^{9}), where a_i denotes the index of the pirate sitting at the round table (n and 1 are neighbours) and b_i the total number of coins that pirate a_i has at the start of the game.
Output
Print 1 if the game is a good game: There is a way to do the splitting so the game ends after finite number of iterations.
Print -1 if the game is a bad game: No matter how You do the splitting the game does not end in finite number of iterations.
Examples
Input
4 2
1 2
2 2
Output
1
Input
6 2
2 3
4 1
Output
1
Input
3 2
1 1
2 2
Output
-1
Note
In the third example the game has no end, because You always only have only one candidate, after whose splitting you end up in the same position as the starting one.
Submitted Solution:
```
n, k = map(int, input().split())
table = {}
for i in range(k):
a, b = map(int, input().split())
table[a] = b
total = 0
for key,value in table.items():
total += value
print(total)
if total % k == 0:
print(1)
else:
print(-1)
```
No
| 94,580 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A famous gang of pirates, Sea Dogs, has come back to their hideout from one of their extravagant plunders. They want to split their treasure fairly amongst themselves, that is why You, their trusted financial advisor, devised a game to help them:
All of them take a sit at their round table, some of them with the golden coins they have just stolen. At each iteration of the game if one of them has equal or more than 2 coins, he is eligible to the splitting and he gives one coin to each pirate sitting next to him. If there are more candidates (pirates with equal or more than 2 coins) then You are the one that chooses which one of them will do the splitting in that iteration. The game ends when there are no more candidates eligible to do the splitting.
Pirates can call it a day, only when the game ends. Since they are beings with a finite amount of time at their disposal, they would prefer if the game that they are playing can end after finite iterations, and if so, they call it a good game. On the other hand, if no matter how You do the splitting, the game cannot end in finite iterations, they call it a bad game. Can You help them figure out before they start playing if the game will be good or bad?
Input
The first line of input contains two integer numbers n and k (1 β€ n β€ 10^{9}, 0 β€ k β€ 2β
10^5), where n denotes total number of pirates and k is the number of pirates that have any coins.
The next k lines of input contain integers a_i and b_i (1 β€ a_i β€ n, 1 β€ b_i β€ 10^{9}), where a_i denotes the index of the pirate sitting at the round table (n and 1 are neighbours) and b_i the total number of coins that pirate a_i has at the start of the game.
Output
Print 1 if the game is a good game: There is a way to do the splitting so the game ends after finite number of iterations.
Print -1 if the game is a bad game: No matter how You do the splitting the game does not end in finite number of iterations.
Examples
Input
4 2
1 2
2 2
Output
1
Input
6 2
2 3
4 1
Output
1
Input
3 2
1 1
2 2
Output
-1
Note
In the third example the game has no end, because You always only have only one candidate, after whose splitting you end up in the same position as the starting one.
Submitted Solution:
```
import sys;input=sys.stdin.readline
N, K = map(int, input().split())
s = 0
X = []
zenb1=1
for _ in range(K):
a, b = map(int, input().split())
X.append((a, b))
s += b
if b != 1:
zenb1 = 0
X.sort(key=lambda x:x[0])
f = 0
ff = 0
T = []
ba = None
for i in range(K):
a, b = X[i]
if a-1 == ba:
T[-1].append((a, b))
else:
T.append([(a, b)])
ba = a
cc = []
for t in T:
if len(t) == 1:
if t[0][1] == 3:
continue
elif t[0][0] == 1 or t[-1][0] == N:
cc.append(t)
else:
ff = 1
break
else:
ft = 1
for x in t[1:-1]:
if x[1] != 1:
ft = 0
break
if (t[0][1] == 2 and t[-1][1] == 2) and ft:
continue
elif t[0][0] == 1 or t[-1][0] == N:
cc.append(t)
else:
ff = 1
break
if len(cc) == 1:
ff = 1
elif len(cc) == 2:
t = cc[0]+cc[1]
ft = 1
for x in t[1:-1]:
if x[1] != 1:
ft = 0
break
if (t[0][1] == 2 and t[-1][1] == 2) and ft:
pass
else:
ff = 1
if s < N:
print(1)
elif s == N:
if zenb1:
print(1)
elif ff:
print(-1)
else:
print(1)
else:
print(-1)
```
No
| 94,581 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A famous gang of pirates, Sea Dogs, has come back to their hideout from one of their extravagant plunders. They want to split their treasure fairly amongst themselves, that is why You, their trusted financial advisor, devised a game to help them:
All of them take a sit at their round table, some of them with the golden coins they have just stolen. At each iteration of the game if one of them has equal or more than 2 coins, he is eligible to the splitting and he gives one coin to each pirate sitting next to him. If there are more candidates (pirates with equal or more than 2 coins) then You are the one that chooses which one of them will do the splitting in that iteration. The game ends when there are no more candidates eligible to do the splitting.
Pirates can call it a day, only when the game ends. Since they are beings with a finite amount of time at their disposal, they would prefer if the game that they are playing can end after finite iterations, and if so, they call it a good game. On the other hand, if no matter how You do the splitting, the game cannot end in finite iterations, they call it a bad game. Can You help them figure out before they start playing if the game will be good or bad?
Input
The first line of input contains two integer numbers n and k (1 β€ n β€ 10^{9}, 0 β€ k β€ 2β
10^5), where n denotes total number of pirates and k is the number of pirates that have any coins.
The next k lines of input contain integers a_i and b_i (1 β€ a_i β€ n, 1 β€ b_i β€ 10^{9}), where a_i denotes the index of the pirate sitting at the round table (n and 1 are neighbours) and b_i the total number of coins that pirate a_i has at the start of the game.
Output
Print 1 if the game is a good game: There is a way to do the splitting so the game ends after finite number of iterations.
Print -1 if the game is a bad game: No matter how You do the splitting the game does not end in finite number of iterations.
Examples
Input
4 2
1 2
2 2
Output
1
Input
6 2
2 3
4 1
Output
1
Input
3 2
1 1
2 2
Output
-1
Note
In the third example the game has no end, because You always only have only one candidate, after whose splitting you end up in the same position as the starting one.
Submitted Solution:
```
from sys import stdin,stdout
from collections import defaultdict as dd
from copy import deepcopy
from math import ceil
input=stdin.readline
mod=10**9+7
for _ in range(1):
n,k=map(int,input().split())
oc=0
r=0
for i in range(k):
a,b= list(map(int,input().split()))
r+=b
if b==1:
oc+=1
if r<n:
print(1)
elif r==n:
if n%2==1:
if oc==n:
print(1)
else:
print(-1)
else:
print(1)
else:
print(-1)
```
No
| 94,582 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A famous gang of pirates, Sea Dogs, has come back to their hideout from one of their extravagant plunders. They want to split their treasure fairly amongst themselves, that is why You, their trusted financial advisor, devised a game to help them:
All of them take a sit at their round table, some of them with the golden coins they have just stolen. At each iteration of the game if one of them has equal or more than 2 coins, he is eligible to the splitting and he gives one coin to each pirate sitting next to him. If there are more candidates (pirates with equal or more than 2 coins) then You are the one that chooses which one of them will do the splitting in that iteration. The game ends when there are no more candidates eligible to do the splitting.
Pirates can call it a day, only when the game ends. Since they are beings with a finite amount of time at their disposal, they would prefer if the game that they are playing can end after finite iterations, and if so, they call it a good game. On the other hand, if no matter how You do the splitting, the game cannot end in finite iterations, they call it a bad game. Can You help them figure out before they start playing if the game will be good or bad?
Input
The first line of input contains two integer numbers n and k (1 β€ n β€ 10^{9}, 0 β€ k β€ 2β
10^5), where n denotes total number of pirates and k is the number of pirates that have any coins.
The next k lines of input contain integers a_i and b_i (1 β€ a_i β€ n, 1 β€ b_i β€ 10^{9}), where a_i denotes the index of the pirate sitting at the round table (n and 1 are neighbours) and b_i the total number of coins that pirate a_i has at the start of the game.
Output
Print 1 if the game is a good game: There is a way to do the splitting so the game ends after finite number of iterations.
Print -1 if the game is a bad game: No matter how You do the splitting the game does not end in finite number of iterations.
Examples
Input
4 2
1 2
2 2
Output
1
Input
6 2
2 3
4 1
Output
1
Input
3 2
1 1
2 2
Output
-1
Note
In the third example the game has no end, because You always only have only one candidate, after whose splitting you end up in the same position as the starting one.
Submitted Solution:
```
a,b = map(int, input().split())
coins = []
count = 0
for i in range(a):
coins.append(0)
for j in range(b):
ind, coin = map(int,input().split())
coins[ind-1] = coin
def solve(count, coins):
print(coins)
if count > 1000:
print(-1)
elif all(x==coins[0] for x in coins):
print(1)
elif all(p < 2 for p in coins):
print(1)
else:
try:
graterIndex = [n for n,i in enumerate(coins) if i>=2 ][0]
if graterIndex == 0:
coins[graterIndex] -=2
coins[len(coins)-1] +=1
coins[1] +=1
count +=1
solve(count, coins)
elif graterIndex == len(coins)-1:
coins[graterIndex] -=2
coins[0] +=1
coins[len(coins)-2] +=1
count +=1
solve(count, coins)
else:
coins[graterIndex] -=2
coins[graterIndex-1] +=1
coins[graterIndex+1] +=1
count +=1
solve(count, coins)
except:
if all(x==coins[0] for x in coins):
print(1)
else:
print(-1)
solve(0, coins)
```
No
| 94,583 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There is a famous olympiad, which has more than a hundred participants. The Olympiad consists of two stages: the elimination stage, and the final stage. At least a hundred participants will advance to the final stage. The elimination stage in turn consists of two contests.
A result of the elimination stage is the total score in two contests, but, unfortunately, the jury lost the final standings and has only standings for the first and for the second contest separately.
In each contest, the participants are ranked by their point score in non-increasing order. When two participants have a tie (earned the same score), they are ranked by their passport number (in accordance with local regulations, all passport numbers are distinct).
In the first contest, the participant on the 100-th place scored a points. Also, the jury checked all participants from the 1-st to the 100-th place (inclusive) in the first contest and found out that all of them have at least b points in the second contest.
Similarly, for the second contest, the participant on the 100-th place has c points. And the jury checked that all the participants from the 1-st to the 100-th place (inclusive) have at least d points in the first contest.
After two contests, all participants are ranked by their total score in two contests in non-increasing order. When participants have the same total score, tie-breaking with passport numbers is used. The cutoff score to qualify to the final stage is the total score of the participant on the 100-th place.
Given integers a, b, c, d, please help the jury determine the smallest possible value of the cutoff score.
Input
You need to process t test cases.
The first line contains an integer t (1 β€ t β€ 3025) β the number of test cases. Then descriptions of t test cases follow.
The first line of each test case contains four integers a, b, c, d (0 β€ a,\,b,\,c,\,d β€ 9; d β€ a; b β€ c).
One can show that for any test case satisfying the constraints above, there is at least one olympiad scenario possible.
Output
For each test case print a single integer β the smallest possible cutoff score in some olympiad scenario satisfying the given information.
Example
Input
2
1 2 2 1
4 8 9 2
Output
3
12
Note
For the first test case, consider the following olympiad scenario: there are 101 participants in the elimination stage, each having 1 point for the first contest and 2 points for the second contest. Hence the total score of the participant on the 100-th place is 3.
For the second test case, consider the following olympiad scenario:
* there are 50 participants with points 5 and 9 for the first and second contest respectively;
* 50 participants with points 4 and 8 for the first and second contest respectively;
* and 50 participants with points 2 and 9 for the first and second contest respectively.
Hence the total point score of the participant on the 100-th place is 12.
Tags: greedy, math
Correct Solution:
```
import sys
from math import *
from fractions import *
def main():
#n = int(sys.stdin.readline().strip())
#p, q = map(int, sys.stdin.readline().split())
#q = list(map(int, sys.stdin.readline().split()))
a,b,c,d= map(int, sys.stdin.readline().split())
print(max(a + b, c + d))
for i in range(int(input()) - 1):
main()
main()
```
| 94,584 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There is a famous olympiad, which has more than a hundred participants. The Olympiad consists of two stages: the elimination stage, and the final stage. At least a hundred participants will advance to the final stage. The elimination stage in turn consists of two contests.
A result of the elimination stage is the total score in two contests, but, unfortunately, the jury lost the final standings and has only standings for the first and for the second contest separately.
In each contest, the participants are ranked by their point score in non-increasing order. When two participants have a tie (earned the same score), they are ranked by their passport number (in accordance with local regulations, all passport numbers are distinct).
In the first contest, the participant on the 100-th place scored a points. Also, the jury checked all participants from the 1-st to the 100-th place (inclusive) in the first contest and found out that all of them have at least b points in the second contest.
Similarly, for the second contest, the participant on the 100-th place has c points. And the jury checked that all the participants from the 1-st to the 100-th place (inclusive) have at least d points in the first contest.
After two contests, all participants are ranked by their total score in two contests in non-increasing order. When participants have the same total score, tie-breaking with passport numbers is used. The cutoff score to qualify to the final stage is the total score of the participant on the 100-th place.
Given integers a, b, c, d, please help the jury determine the smallest possible value of the cutoff score.
Input
You need to process t test cases.
The first line contains an integer t (1 β€ t β€ 3025) β the number of test cases. Then descriptions of t test cases follow.
The first line of each test case contains four integers a, b, c, d (0 β€ a,\,b,\,c,\,d β€ 9; d β€ a; b β€ c).
One can show that for any test case satisfying the constraints above, there is at least one olympiad scenario possible.
Output
For each test case print a single integer β the smallest possible cutoff score in some olympiad scenario satisfying the given information.
Example
Input
2
1 2 2 1
4 8 9 2
Output
3
12
Note
For the first test case, consider the following olympiad scenario: there are 101 participants in the elimination stage, each having 1 point for the first contest and 2 points for the second contest. Hence the total score of the participant on the 100-th place is 3.
For the second test case, consider the following olympiad scenario:
* there are 50 participants with points 5 and 9 for the first and second contest respectively;
* 50 participants with points 4 and 8 for the first and second contest respectively;
* and 50 participants with points 2 and 9 for the first and second contest respectively.
Hence the total point score of the participant on the 100-th place is 12.
Tags: greedy, math
Correct Solution:
```
if __name__ == "__main__":
t = int(input())
while t:
a,b,c,d = list(map(int, input().split()))
print(max( min((a+b),(a+c)),(c+d) ))
t = t - 1
```
| 94,585 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There is a famous olympiad, which has more than a hundred participants. The Olympiad consists of two stages: the elimination stage, and the final stage. At least a hundred participants will advance to the final stage. The elimination stage in turn consists of two contests.
A result of the elimination stage is the total score in two contests, but, unfortunately, the jury lost the final standings and has only standings for the first and for the second contest separately.
In each contest, the participants are ranked by their point score in non-increasing order. When two participants have a tie (earned the same score), they are ranked by their passport number (in accordance with local regulations, all passport numbers are distinct).
In the first contest, the participant on the 100-th place scored a points. Also, the jury checked all participants from the 1-st to the 100-th place (inclusive) in the first contest and found out that all of them have at least b points in the second contest.
Similarly, for the second contest, the participant on the 100-th place has c points. And the jury checked that all the participants from the 1-st to the 100-th place (inclusive) have at least d points in the first contest.
After two contests, all participants are ranked by their total score in two contests in non-increasing order. When participants have the same total score, tie-breaking with passport numbers is used. The cutoff score to qualify to the final stage is the total score of the participant on the 100-th place.
Given integers a, b, c, d, please help the jury determine the smallest possible value of the cutoff score.
Input
You need to process t test cases.
The first line contains an integer t (1 β€ t β€ 3025) β the number of test cases. Then descriptions of t test cases follow.
The first line of each test case contains four integers a, b, c, d (0 β€ a,\,b,\,c,\,d β€ 9; d β€ a; b β€ c).
One can show that for any test case satisfying the constraints above, there is at least one olympiad scenario possible.
Output
For each test case print a single integer β the smallest possible cutoff score in some olympiad scenario satisfying the given information.
Example
Input
2
1 2 2 1
4 8 9 2
Output
3
12
Note
For the first test case, consider the following olympiad scenario: there are 101 participants in the elimination stage, each having 1 point for the first contest and 2 points for the second contest. Hence the total score of the participant on the 100-th place is 3.
For the second test case, consider the following olympiad scenario:
* there are 50 participants with points 5 and 9 for the first and second contest respectively;
* 50 participants with points 4 and 8 for the first and second contest respectively;
* and 50 participants with points 2 and 9 for the first and second contest respectively.
Hence the total point score of the participant on the 100-th place is 12.
Tags: greedy, math
Correct Solution:
```
for _ in range(int(input())):
a,b,c,d=map(int,input().split())
print(a+b if (a+b)>=(c+d) else c+d)
```
| 94,586 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There is a famous olympiad, which has more than a hundred participants. The Olympiad consists of two stages: the elimination stage, and the final stage. At least a hundred participants will advance to the final stage. The elimination stage in turn consists of two contests.
A result of the elimination stage is the total score in two contests, but, unfortunately, the jury lost the final standings and has only standings for the first and for the second contest separately.
In each contest, the participants are ranked by their point score in non-increasing order. When two participants have a tie (earned the same score), they are ranked by their passport number (in accordance with local regulations, all passport numbers are distinct).
In the first contest, the participant on the 100-th place scored a points. Also, the jury checked all participants from the 1-st to the 100-th place (inclusive) in the first contest and found out that all of them have at least b points in the second contest.
Similarly, for the second contest, the participant on the 100-th place has c points. And the jury checked that all the participants from the 1-st to the 100-th place (inclusive) have at least d points in the first contest.
After two contests, all participants are ranked by their total score in two contests in non-increasing order. When participants have the same total score, tie-breaking with passport numbers is used. The cutoff score to qualify to the final stage is the total score of the participant on the 100-th place.
Given integers a, b, c, d, please help the jury determine the smallest possible value of the cutoff score.
Input
You need to process t test cases.
The first line contains an integer t (1 β€ t β€ 3025) β the number of test cases. Then descriptions of t test cases follow.
The first line of each test case contains four integers a, b, c, d (0 β€ a,\,b,\,c,\,d β€ 9; d β€ a; b β€ c).
One can show that for any test case satisfying the constraints above, there is at least one olympiad scenario possible.
Output
For each test case print a single integer β the smallest possible cutoff score in some olympiad scenario satisfying the given information.
Example
Input
2
1 2 2 1
4 8 9 2
Output
3
12
Note
For the first test case, consider the following olympiad scenario: there are 101 participants in the elimination stage, each having 1 point for the first contest and 2 points for the second contest. Hence the total score of the participant on the 100-th place is 3.
For the second test case, consider the following olympiad scenario:
* there are 50 participants with points 5 and 9 for the first and second contest respectively;
* 50 participants with points 4 and 8 for the first and second contest respectively;
* and 50 participants with points 2 and 9 for the first and second contest respectively.
Hence the total point score of the participant on the 100-th place is 12.
Tags: greedy, math
Correct Solution:
```
def zip_sorted(a,b):
# sorted by a
a,b = zip(*sorted(zip(a,b)))
# sorted by b
sorted(zip(a, b), key=lambda x: x[1])
return a,b
import sys
input = sys.stdin.readline
I = lambda : list(map(int,input().split()))
S = lambda : list(map(str,input().split()))
t,=I()
for t1 in range(t):
a,b,c,d = I()
print(max((a+b),(c+d)))
```
| 94,587 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There is a famous olympiad, which has more than a hundred participants. The Olympiad consists of two stages: the elimination stage, and the final stage. At least a hundred participants will advance to the final stage. The elimination stage in turn consists of two contests.
A result of the elimination stage is the total score in two contests, but, unfortunately, the jury lost the final standings and has only standings for the first and for the second contest separately.
In each contest, the participants are ranked by their point score in non-increasing order. When two participants have a tie (earned the same score), they are ranked by their passport number (in accordance with local regulations, all passport numbers are distinct).
In the first contest, the participant on the 100-th place scored a points. Also, the jury checked all participants from the 1-st to the 100-th place (inclusive) in the first contest and found out that all of them have at least b points in the second contest.
Similarly, for the second contest, the participant on the 100-th place has c points. And the jury checked that all the participants from the 1-st to the 100-th place (inclusive) have at least d points in the first contest.
After two contests, all participants are ranked by their total score in two contests in non-increasing order. When participants have the same total score, tie-breaking with passport numbers is used. The cutoff score to qualify to the final stage is the total score of the participant on the 100-th place.
Given integers a, b, c, d, please help the jury determine the smallest possible value of the cutoff score.
Input
You need to process t test cases.
The first line contains an integer t (1 β€ t β€ 3025) β the number of test cases. Then descriptions of t test cases follow.
The first line of each test case contains four integers a, b, c, d (0 β€ a,\,b,\,c,\,d β€ 9; d β€ a; b β€ c).
One can show that for any test case satisfying the constraints above, there is at least one olympiad scenario possible.
Output
For each test case print a single integer β the smallest possible cutoff score in some olympiad scenario satisfying the given information.
Example
Input
2
1 2 2 1
4 8 9 2
Output
3
12
Note
For the first test case, consider the following olympiad scenario: there are 101 participants in the elimination stage, each having 1 point for the first contest and 2 points for the second contest. Hence the total score of the participant on the 100-th place is 3.
For the second test case, consider the following olympiad scenario:
* there are 50 participants with points 5 and 9 for the first and second contest respectively;
* 50 participants with points 4 and 8 for the first and second contest respectively;
* and 50 participants with points 2 and 9 for the first and second contest respectively.
Hence the total point score of the participant on the 100-th place is 12.
Tags: greedy, math
Correct Solution:
```
try:
for _ in range(int(input())):
a,b,c,d=map(int,input().split())
print(max((a+b),(c+d)))
except:
pass
```
| 94,588 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There is a famous olympiad, which has more than a hundred participants. The Olympiad consists of two stages: the elimination stage, and the final stage. At least a hundred participants will advance to the final stage. The elimination stage in turn consists of two contests.
A result of the elimination stage is the total score in two contests, but, unfortunately, the jury lost the final standings and has only standings for the first and for the second contest separately.
In each contest, the participants are ranked by their point score in non-increasing order. When two participants have a tie (earned the same score), they are ranked by their passport number (in accordance with local regulations, all passport numbers are distinct).
In the first contest, the participant on the 100-th place scored a points. Also, the jury checked all participants from the 1-st to the 100-th place (inclusive) in the first contest and found out that all of them have at least b points in the second contest.
Similarly, for the second contest, the participant on the 100-th place has c points. And the jury checked that all the participants from the 1-st to the 100-th place (inclusive) have at least d points in the first contest.
After two contests, all participants are ranked by their total score in two contests in non-increasing order. When participants have the same total score, tie-breaking with passport numbers is used. The cutoff score to qualify to the final stage is the total score of the participant on the 100-th place.
Given integers a, b, c, d, please help the jury determine the smallest possible value of the cutoff score.
Input
You need to process t test cases.
The first line contains an integer t (1 β€ t β€ 3025) β the number of test cases. Then descriptions of t test cases follow.
The first line of each test case contains four integers a, b, c, d (0 β€ a,\,b,\,c,\,d β€ 9; d β€ a; b β€ c).
One can show that for any test case satisfying the constraints above, there is at least one olympiad scenario possible.
Output
For each test case print a single integer β the smallest possible cutoff score in some olympiad scenario satisfying the given information.
Example
Input
2
1 2 2 1
4 8 9 2
Output
3
12
Note
For the first test case, consider the following olympiad scenario: there are 101 participants in the elimination stage, each having 1 point for the first contest and 2 points for the second contest. Hence the total score of the participant on the 100-th place is 3.
For the second test case, consider the following olympiad scenario:
* there are 50 participants with points 5 and 9 for the first and second contest respectively;
* 50 participants with points 4 and 8 for the first and second contest respectively;
* and 50 participants with points 2 and 9 for the first and second contest respectively.
Hence the total point score of the participant on the 100-th place is 12.
Tags: greedy, math
Correct Solution:
```
def main():
t = int(input())
for i in range(t):
a, b, c, d = map(int, input().split())
print(max(a + b, c + d))
main()
```
| 94,589 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There is a famous olympiad, which has more than a hundred participants. The Olympiad consists of two stages: the elimination stage, and the final stage. At least a hundred participants will advance to the final stage. The elimination stage in turn consists of two contests.
A result of the elimination stage is the total score in two contests, but, unfortunately, the jury lost the final standings and has only standings for the first and for the second contest separately.
In each contest, the participants are ranked by their point score in non-increasing order. When two participants have a tie (earned the same score), they are ranked by their passport number (in accordance with local regulations, all passport numbers are distinct).
In the first contest, the participant on the 100-th place scored a points. Also, the jury checked all participants from the 1-st to the 100-th place (inclusive) in the first contest and found out that all of them have at least b points in the second contest.
Similarly, for the second contest, the participant on the 100-th place has c points. And the jury checked that all the participants from the 1-st to the 100-th place (inclusive) have at least d points in the first contest.
After two contests, all participants are ranked by their total score in two contests in non-increasing order. When participants have the same total score, tie-breaking with passport numbers is used. The cutoff score to qualify to the final stage is the total score of the participant on the 100-th place.
Given integers a, b, c, d, please help the jury determine the smallest possible value of the cutoff score.
Input
You need to process t test cases.
The first line contains an integer t (1 β€ t β€ 3025) β the number of test cases. Then descriptions of t test cases follow.
The first line of each test case contains four integers a, b, c, d (0 β€ a,\,b,\,c,\,d β€ 9; d β€ a; b β€ c).
One can show that for any test case satisfying the constraints above, there is at least one olympiad scenario possible.
Output
For each test case print a single integer β the smallest possible cutoff score in some olympiad scenario satisfying the given information.
Example
Input
2
1 2 2 1
4 8 9 2
Output
3
12
Note
For the first test case, consider the following olympiad scenario: there are 101 participants in the elimination stage, each having 1 point for the first contest and 2 points for the second contest. Hence the total score of the participant on the 100-th place is 3.
For the second test case, consider the following olympiad scenario:
* there are 50 participants with points 5 and 9 for the first and second contest respectively;
* 50 participants with points 4 and 8 for the first and second contest respectively;
* and 50 participants with points 2 and 9 for the first and second contest respectively.
Hence the total point score of the participant on the 100-th place is 12.
Tags: greedy, math
Correct Solution:
```
import sys, math
import io, os
#data = io.BytesIO(os.read(0,os.fstat(0).st_size)).readline
from bisect import bisect_left as bl, bisect_right as br, insort
from heapq import heapify, heappush, heappop
from collections import defaultdict as dd, deque, Counter
from itertools import permutations,combinations
def data(): return sys.stdin.readline().strip()
def mdata(): return list(map(int, data().split()))
def outl(var) : sys.stdout.write(' '.join(map(str, var))+'\n')
def out(var) : sys.stdout.write(str(var)+'\n')
#from decimal import Decimal
#from fractions import Fraction
sys.setrecursionlimit(100000)
INF = float('inf')
mod = 998244353
for t in range(int(data())):
a, b,c ,d=mdata()
out(max(a+b,c+d))
```
| 94,590 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There is a famous olympiad, which has more than a hundred participants. The Olympiad consists of two stages: the elimination stage, and the final stage. At least a hundred participants will advance to the final stage. The elimination stage in turn consists of two contests.
A result of the elimination stage is the total score in two contests, but, unfortunately, the jury lost the final standings and has only standings for the first and for the second contest separately.
In each contest, the participants are ranked by their point score in non-increasing order. When two participants have a tie (earned the same score), they are ranked by their passport number (in accordance with local regulations, all passport numbers are distinct).
In the first contest, the participant on the 100-th place scored a points. Also, the jury checked all participants from the 1-st to the 100-th place (inclusive) in the first contest and found out that all of them have at least b points in the second contest.
Similarly, for the second contest, the participant on the 100-th place has c points. And the jury checked that all the participants from the 1-st to the 100-th place (inclusive) have at least d points in the first contest.
After two contests, all participants are ranked by their total score in two contests in non-increasing order. When participants have the same total score, tie-breaking with passport numbers is used. The cutoff score to qualify to the final stage is the total score of the participant on the 100-th place.
Given integers a, b, c, d, please help the jury determine the smallest possible value of the cutoff score.
Input
You need to process t test cases.
The first line contains an integer t (1 β€ t β€ 3025) β the number of test cases. Then descriptions of t test cases follow.
The first line of each test case contains four integers a, b, c, d (0 β€ a,\,b,\,c,\,d β€ 9; d β€ a; b β€ c).
One can show that for any test case satisfying the constraints above, there is at least one olympiad scenario possible.
Output
For each test case print a single integer β the smallest possible cutoff score in some olympiad scenario satisfying the given information.
Example
Input
2
1 2 2 1
4 8 9 2
Output
3
12
Note
For the first test case, consider the following olympiad scenario: there are 101 participants in the elimination stage, each having 1 point for the first contest and 2 points for the second contest. Hence the total score of the participant on the 100-th place is 3.
For the second test case, consider the following olympiad scenario:
* there are 50 participants with points 5 and 9 for the first and second contest respectively;
* 50 participants with points 4 and 8 for the first and second contest respectively;
* and 50 participants with points 2 and 9 for the first and second contest respectively.
Hence the total point score of the participant on the 100-th place is 12.
Tags: greedy, math
Correct Solution:
```
t = int(input())
while t > 0:
a, b, c, d = [int(x) for x in input().split()]
ans = max(a + b, c + d)
print(ans)
t -= 1
```
| 94,591 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a famous olympiad, which has more than a hundred participants. The Olympiad consists of two stages: the elimination stage, and the final stage. At least a hundred participants will advance to the final stage. The elimination stage in turn consists of two contests.
A result of the elimination stage is the total score in two contests, but, unfortunately, the jury lost the final standings and has only standings for the first and for the second contest separately.
In each contest, the participants are ranked by their point score in non-increasing order. When two participants have a tie (earned the same score), they are ranked by their passport number (in accordance with local regulations, all passport numbers are distinct).
In the first contest, the participant on the 100-th place scored a points. Also, the jury checked all participants from the 1-st to the 100-th place (inclusive) in the first contest and found out that all of them have at least b points in the second contest.
Similarly, for the second contest, the participant on the 100-th place has c points. And the jury checked that all the participants from the 1-st to the 100-th place (inclusive) have at least d points in the first contest.
After two contests, all participants are ranked by their total score in two contests in non-increasing order. When participants have the same total score, tie-breaking with passport numbers is used. The cutoff score to qualify to the final stage is the total score of the participant on the 100-th place.
Given integers a, b, c, d, please help the jury determine the smallest possible value of the cutoff score.
Input
You need to process t test cases.
The first line contains an integer t (1 β€ t β€ 3025) β the number of test cases. Then descriptions of t test cases follow.
The first line of each test case contains four integers a, b, c, d (0 β€ a,\,b,\,c,\,d β€ 9; d β€ a; b β€ c).
One can show that for any test case satisfying the constraints above, there is at least one olympiad scenario possible.
Output
For each test case print a single integer β the smallest possible cutoff score in some olympiad scenario satisfying the given information.
Example
Input
2
1 2 2 1
4 8 9 2
Output
3
12
Note
For the first test case, consider the following olympiad scenario: there are 101 participants in the elimination stage, each having 1 point for the first contest and 2 points for the second contest. Hence the total score of the participant on the 100-th place is 3.
For the second test case, consider the following olympiad scenario:
* there are 50 participants with points 5 and 9 for the first and second contest respectively;
* 50 participants with points 4 and 8 for the first and second contest respectively;
* and 50 participants with points 2 and 9 for the first and second contest respectively.
Hence the total point score of the participant on the 100-th place is 12.
Submitted Solution:
```
for i in range(int(input())):
a,b,c,d=map(int,input().split())
print(max((a+b),(c+d)))
```
Yes
| 94,592 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a famous olympiad, which has more than a hundred participants. The Olympiad consists of two stages: the elimination stage, and the final stage. At least a hundred participants will advance to the final stage. The elimination stage in turn consists of two contests.
A result of the elimination stage is the total score in two contests, but, unfortunately, the jury lost the final standings and has only standings for the first and for the second contest separately.
In each contest, the participants are ranked by their point score in non-increasing order. When two participants have a tie (earned the same score), they are ranked by their passport number (in accordance with local regulations, all passport numbers are distinct).
In the first contest, the participant on the 100-th place scored a points. Also, the jury checked all participants from the 1-st to the 100-th place (inclusive) in the first contest and found out that all of them have at least b points in the second contest.
Similarly, for the second contest, the participant on the 100-th place has c points. And the jury checked that all the participants from the 1-st to the 100-th place (inclusive) have at least d points in the first contest.
After two contests, all participants are ranked by their total score in two contests in non-increasing order. When participants have the same total score, tie-breaking with passport numbers is used. The cutoff score to qualify to the final stage is the total score of the participant on the 100-th place.
Given integers a, b, c, d, please help the jury determine the smallest possible value of the cutoff score.
Input
You need to process t test cases.
The first line contains an integer t (1 β€ t β€ 3025) β the number of test cases. Then descriptions of t test cases follow.
The first line of each test case contains four integers a, b, c, d (0 β€ a,\,b,\,c,\,d β€ 9; d β€ a; b β€ c).
One can show that for any test case satisfying the constraints above, there is at least one olympiad scenario possible.
Output
For each test case print a single integer β the smallest possible cutoff score in some olympiad scenario satisfying the given information.
Example
Input
2
1 2 2 1
4 8 9 2
Output
3
12
Note
For the first test case, consider the following olympiad scenario: there are 101 participants in the elimination stage, each having 1 point for the first contest and 2 points for the second contest. Hence the total score of the participant on the 100-th place is 3.
For the second test case, consider the following olympiad scenario:
* there are 50 participants with points 5 and 9 for the first and second contest respectively;
* 50 participants with points 4 and 8 for the first and second contest respectively;
* and 50 participants with points 2 and 9 for the first and second contest respectively.
Hence the total point score of the participant on the 100-th place is 12.
Submitted Solution:
```
#(γ£ββ‘β)γ£ β₯ GLHF β₯
import os #(γ£ββ‘β)γ£
import sys #(γ£ββ‘β)γ£
from io import BytesIO, IOBase #(γ£ββ‘β)γ£
def main(): #(γ£ββ‘β)γ£
for tc in range(int(input())):
line = input().split()
a100, b, b100, a = int(line[0]), int(line[1]), int(line[2]), int(line[3])
print(max(a100+b, b100+a))
BUFSIZE = 8192 #(γ£ββ‘β)γ£
class FastIO(IOBase): #(γ£ββ‘β)γ£
newlines = 0 #(γ£ββ‘β)γ£
def __init__(self, file): #(γ£ββ‘β)γ£
self._fd = file.fileno() #(γ£ββ‘β)γ£
self.buffer = BytesIO() #(γ£ββ‘β)γ£
self.writable = "x" in file.mode or "r" not in file.mode #(γ£ββ‘β)γ£
self.write = self.buffer.write if self.writable else None #(γ£ββ‘β)γ£
def read(self): #(γ£ββ‘β)γ£
while True: #(γ£ββ‘β)γ£
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE)) #(γ£ββ‘β)γ£
if not b: #(γ£ββ‘β)γ£
break #(γ£ββ‘β)γ£
ptr = self.buffer.tell() #(γ£ββ‘β)γ£
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr) #(γ£ββ‘β)γ£
self.newlines = 0 #(γ£ββ‘β)γ£
return self.buffer.read() #(γ£ββ‘β)γ£
def readline(self): #(γ£ββ‘β)γ£
while self.newlines == 0: #(γ£ββ‘β)γ£
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE)) #(γ£ββ‘β)γ£
self.newlines = b.count(b"\n") + (not b) #(γ£ββ‘β)γ£
ptr = self.buffer.tell() #(γ£ββ‘β)γ£
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr) #(γ£ββ‘β)γ£
self.newlines -= 1 #(γ£ββ‘β)γ£
return self.buffer.readline() #(γ£ββ‘β)γ£
def flush(self): #(γ£ββ‘β)γ£
if self.writable: #(γ£ββ‘β)γ£
os.write(self._fd, self.buffer.getvalue()) #(γ£ββ‘β)γ£
self.buffer.truncate(0), self.buffer.seek(0) #(γ£ββ‘β)γ£
class IOWrapper(IOBase): #(γ£ββ‘β)γ£
def __init__(self, file): #(γ£ββ‘β)γ£
self.buffer = FastIO(file) #(γ£ββ‘β)γ£
self.flush = self.buffer.flush #(γ£ββ‘β)γ£
self.writable = self.buffer.writable #(γ£ββ‘β)γ£
self.write = lambda s: self.buffer.write(s.encode("ascii")) #(γ£ββ‘β)γ£
self.read = lambda: self.buffer.read().decode("ascii") #(γ£ββ‘β)γ£
self.readline = lambda: self.buffer.readline().decode("ascii") #(γ£ββ‘β)γ£
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout) #(γ£ββ‘β)γ£
input = lambda: sys.stdin.readline().rstrip("\r\n") #(γ£ββ‘β)γ£
if __name__ == "__main__": #(γ£ββ‘β)γ£
main() #(γ£ββ‘β)γ£
#ββββββββββββββββββββββββββββββββββββ
#ββββββββββββββββββββββββββββββββββββ
#ββββββββββββββββββββββββββββββββββββ
#ββββββββββββββββββββββββββββββββββββ
#ββββββββββββββββββββββββββββββββββββ
#ββββββββββββββββββββββββββββββββββββ
```
Yes
| 94,593 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a famous olympiad, which has more than a hundred participants. The Olympiad consists of two stages: the elimination stage, and the final stage. At least a hundred participants will advance to the final stage. The elimination stage in turn consists of two contests.
A result of the elimination stage is the total score in two contests, but, unfortunately, the jury lost the final standings and has only standings for the first and for the second contest separately.
In each contest, the participants are ranked by their point score in non-increasing order. When two participants have a tie (earned the same score), they are ranked by their passport number (in accordance with local regulations, all passport numbers are distinct).
In the first contest, the participant on the 100-th place scored a points. Also, the jury checked all participants from the 1-st to the 100-th place (inclusive) in the first contest and found out that all of them have at least b points in the second contest.
Similarly, for the second contest, the participant on the 100-th place has c points. And the jury checked that all the participants from the 1-st to the 100-th place (inclusive) have at least d points in the first contest.
After two contests, all participants are ranked by their total score in two contests in non-increasing order. When participants have the same total score, tie-breaking with passport numbers is used. The cutoff score to qualify to the final stage is the total score of the participant on the 100-th place.
Given integers a, b, c, d, please help the jury determine the smallest possible value of the cutoff score.
Input
You need to process t test cases.
The first line contains an integer t (1 β€ t β€ 3025) β the number of test cases. Then descriptions of t test cases follow.
The first line of each test case contains four integers a, b, c, d (0 β€ a,\,b,\,c,\,d β€ 9; d β€ a; b β€ c).
One can show that for any test case satisfying the constraints above, there is at least one olympiad scenario possible.
Output
For each test case print a single integer β the smallest possible cutoff score in some olympiad scenario satisfying the given information.
Example
Input
2
1 2 2 1
4 8 9 2
Output
3
12
Note
For the first test case, consider the following olympiad scenario: there are 101 participants in the elimination stage, each having 1 point for the first contest and 2 points for the second contest. Hence the total score of the participant on the 100-th place is 3.
For the second test case, consider the following olympiad scenario:
* there are 50 participants with points 5 and 9 for the first and second contest respectively;
* 50 participants with points 4 and 8 for the first and second contest respectively;
* and 50 participants with points 2 and 9 for the first and second contest respectively.
Hence the total point score of the participant on the 100-th place is 12.
Submitted Solution:
```
for _ in range(int(input())):
a, b, c, d = map(int, input().split())
print(max(a + b, c + d)) if a >= d else print(a + b)
```
Yes
| 94,594 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a famous olympiad, which has more than a hundred participants. The Olympiad consists of two stages: the elimination stage, and the final stage. At least a hundred participants will advance to the final stage. The elimination stage in turn consists of two contests.
A result of the elimination stage is the total score in two contests, but, unfortunately, the jury lost the final standings and has only standings for the first and for the second contest separately.
In each contest, the participants are ranked by their point score in non-increasing order. When two participants have a tie (earned the same score), they are ranked by their passport number (in accordance with local regulations, all passport numbers are distinct).
In the first contest, the participant on the 100-th place scored a points. Also, the jury checked all participants from the 1-st to the 100-th place (inclusive) in the first contest and found out that all of them have at least b points in the second contest.
Similarly, for the second contest, the participant on the 100-th place has c points. And the jury checked that all the participants from the 1-st to the 100-th place (inclusive) have at least d points in the first contest.
After two contests, all participants are ranked by their total score in two contests in non-increasing order. When participants have the same total score, tie-breaking with passport numbers is used. The cutoff score to qualify to the final stage is the total score of the participant on the 100-th place.
Given integers a, b, c, d, please help the jury determine the smallest possible value of the cutoff score.
Input
You need to process t test cases.
The first line contains an integer t (1 β€ t β€ 3025) β the number of test cases. Then descriptions of t test cases follow.
The first line of each test case contains four integers a, b, c, d (0 β€ a,\,b,\,c,\,d β€ 9; d β€ a; b β€ c).
One can show that for any test case satisfying the constraints above, there is at least one olympiad scenario possible.
Output
For each test case print a single integer β the smallest possible cutoff score in some olympiad scenario satisfying the given information.
Example
Input
2
1 2 2 1
4 8 9 2
Output
3
12
Note
For the first test case, consider the following olympiad scenario: there are 101 participants in the elimination stage, each having 1 point for the first contest and 2 points for the second contest. Hence the total score of the participant on the 100-th place is 3.
For the second test case, consider the following olympiad scenario:
* there are 50 participants with points 5 and 9 for the first and second contest respectively;
* 50 participants with points 4 and 8 for the first and second contest respectively;
* and 50 participants with points 2 and 9 for the first and second contest respectively.
Hence the total point score of the participant on the 100-th place is 12.
Submitted Solution:
```
t = int(input())
for _ in range(t):
a, b, c, d = [int(x) for x in input().split()]
print(max(a+b, c+d))
```
Yes
| 94,595 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a famous olympiad, which has more than a hundred participants. The Olympiad consists of two stages: the elimination stage, and the final stage. At least a hundred participants will advance to the final stage. The elimination stage in turn consists of two contests.
A result of the elimination stage is the total score in two contests, but, unfortunately, the jury lost the final standings and has only standings for the first and for the second contest separately.
In each contest, the participants are ranked by their point score in non-increasing order. When two participants have a tie (earned the same score), they are ranked by their passport number (in accordance with local regulations, all passport numbers are distinct).
In the first contest, the participant on the 100-th place scored a points. Also, the jury checked all participants from the 1-st to the 100-th place (inclusive) in the first contest and found out that all of them have at least b points in the second contest.
Similarly, for the second contest, the participant on the 100-th place has c points. And the jury checked that all the participants from the 1-st to the 100-th place (inclusive) have at least d points in the first contest.
After two contests, all participants are ranked by their total score in two contests in non-increasing order. When participants have the same total score, tie-breaking with passport numbers is used. The cutoff score to qualify to the final stage is the total score of the participant on the 100-th place.
Given integers a, b, c, d, please help the jury determine the smallest possible value of the cutoff score.
Input
You need to process t test cases.
The first line contains an integer t (1 β€ t β€ 3025) β the number of test cases. Then descriptions of t test cases follow.
The first line of each test case contains four integers a, b, c, d (0 β€ a,\,b,\,c,\,d β€ 9; d β€ a; b β€ c).
One can show that for any test case satisfying the constraints above, there is at least one olympiad scenario possible.
Output
For each test case print a single integer β the smallest possible cutoff score in some olympiad scenario satisfying the given information.
Example
Input
2
1 2 2 1
4 8 9 2
Output
3
12
Note
For the first test case, consider the following olympiad scenario: there are 101 participants in the elimination stage, each having 1 point for the first contest and 2 points for the second contest. Hence the total score of the participant on the 100-th place is 3.
For the second test case, consider the following olympiad scenario:
* there are 50 participants with points 5 and 9 for the first and second contest respectively;
* 50 participants with points 4 and 8 for the first and second contest respectively;
* and 50 participants with points 2 and 9 for the first and second contest respectively.
Hence the total point score of the participant on the 100-th place is 12.
Submitted Solution:
```
t=int(input())
for i in range(t):
a,b,c,d=map(int,input().strip().split())
an=a+c
if d<a:
an=min(an,d+1+c)
print(an)
```
No
| 94,596 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a famous olympiad, which has more than a hundred participants. The Olympiad consists of two stages: the elimination stage, and the final stage. At least a hundred participants will advance to the final stage. The elimination stage in turn consists of two contests.
A result of the elimination stage is the total score in two contests, but, unfortunately, the jury lost the final standings and has only standings for the first and for the second contest separately.
In each contest, the participants are ranked by their point score in non-increasing order. When two participants have a tie (earned the same score), they are ranked by their passport number (in accordance with local regulations, all passport numbers are distinct).
In the first contest, the participant on the 100-th place scored a points. Also, the jury checked all participants from the 1-st to the 100-th place (inclusive) in the first contest and found out that all of them have at least b points in the second contest.
Similarly, for the second contest, the participant on the 100-th place has c points. And the jury checked that all the participants from the 1-st to the 100-th place (inclusive) have at least d points in the first contest.
After two contests, all participants are ranked by their total score in two contests in non-increasing order. When participants have the same total score, tie-breaking with passport numbers is used. The cutoff score to qualify to the final stage is the total score of the participant on the 100-th place.
Given integers a, b, c, d, please help the jury determine the smallest possible value of the cutoff score.
Input
You need to process t test cases.
The first line contains an integer t (1 β€ t β€ 3025) β the number of test cases. Then descriptions of t test cases follow.
The first line of each test case contains four integers a, b, c, d (0 β€ a,\,b,\,c,\,d β€ 9; d β€ a; b β€ c).
One can show that for any test case satisfying the constraints above, there is at least one olympiad scenario possible.
Output
For each test case print a single integer β the smallest possible cutoff score in some olympiad scenario satisfying the given information.
Example
Input
2
1 2 2 1
4 8 9 2
Output
3
12
Note
For the first test case, consider the following olympiad scenario: there are 101 participants in the elimination stage, each having 1 point for the first contest and 2 points for the second contest. Hence the total score of the participant on the 100-th place is 3.
For the second test case, consider the following olympiad scenario:
* there are 50 participants with points 5 and 9 for the first and second contest respectively;
* 50 participants with points 4 and 8 for the first and second contest respectively;
* and 50 participants with points 2 and 9 for the first and second contest respectively.
Hence the total point score of the participant on the 100-th place is 12.
Submitted Solution:
```
for _ in range(int(input())):
a, b, c, d = map(int, input().split())
if a > d:
print(a + b)
else:
print(c + d)
```
No
| 94,597 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a famous olympiad, which has more than a hundred participants. The Olympiad consists of two stages: the elimination stage, and the final stage. At least a hundred participants will advance to the final stage. The elimination stage in turn consists of two contests.
A result of the elimination stage is the total score in two contests, but, unfortunately, the jury lost the final standings and has only standings for the first and for the second contest separately.
In each contest, the participants are ranked by their point score in non-increasing order. When two participants have a tie (earned the same score), they are ranked by their passport number (in accordance with local regulations, all passport numbers are distinct).
In the first contest, the participant on the 100-th place scored a points. Also, the jury checked all participants from the 1-st to the 100-th place (inclusive) in the first contest and found out that all of them have at least b points in the second contest.
Similarly, for the second contest, the participant on the 100-th place has c points. And the jury checked that all the participants from the 1-st to the 100-th place (inclusive) have at least d points in the first contest.
After two contests, all participants are ranked by their total score in two contests in non-increasing order. When participants have the same total score, tie-breaking with passport numbers is used. The cutoff score to qualify to the final stage is the total score of the participant on the 100-th place.
Given integers a, b, c, d, please help the jury determine the smallest possible value of the cutoff score.
Input
You need to process t test cases.
The first line contains an integer t (1 β€ t β€ 3025) β the number of test cases. Then descriptions of t test cases follow.
The first line of each test case contains four integers a, b, c, d (0 β€ a,\,b,\,c,\,d β€ 9; d β€ a; b β€ c).
One can show that for any test case satisfying the constraints above, there is at least one olympiad scenario possible.
Output
For each test case print a single integer β the smallest possible cutoff score in some olympiad scenario satisfying the given information.
Example
Input
2
1 2 2 1
4 8 9 2
Output
3
12
Note
For the first test case, consider the following olympiad scenario: there are 101 participants in the elimination stage, each having 1 point for the first contest and 2 points for the second contest. Hence the total score of the participant on the 100-th place is 3.
For the second test case, consider the following olympiad scenario:
* there are 50 participants with points 5 and 9 for the first and second contest respectively;
* 50 participants with points 4 and 8 for the first and second contest respectively;
* and 50 participants with points 2 and 9 for the first and second contest respectively.
Hence the total point score of the participant on the 100-th place is 12.
Submitted Solution:
```
import math
def power(x,y):
res=1
while y>0:
if y&1:
res=(res*x)
y=y>>1
x=(x*x)
return res
t=int(input())
for tt in range(t):
a,b,c,d=map(int,input().split())
print(max(a,d)+min(b,c))
```
No
| 94,598 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a famous olympiad, which has more than a hundred participants. The Olympiad consists of two stages: the elimination stage, and the final stage. At least a hundred participants will advance to the final stage. The elimination stage in turn consists of two contests.
A result of the elimination stage is the total score in two contests, but, unfortunately, the jury lost the final standings and has only standings for the first and for the second contest separately.
In each contest, the participants are ranked by their point score in non-increasing order. When two participants have a tie (earned the same score), they are ranked by their passport number (in accordance with local regulations, all passport numbers are distinct).
In the first contest, the participant on the 100-th place scored a points. Also, the jury checked all participants from the 1-st to the 100-th place (inclusive) in the first contest and found out that all of them have at least b points in the second contest.
Similarly, for the second contest, the participant on the 100-th place has c points. And the jury checked that all the participants from the 1-st to the 100-th place (inclusive) have at least d points in the first contest.
After two contests, all participants are ranked by their total score in two contests in non-increasing order. When participants have the same total score, tie-breaking with passport numbers is used. The cutoff score to qualify to the final stage is the total score of the participant on the 100-th place.
Given integers a, b, c, d, please help the jury determine the smallest possible value of the cutoff score.
Input
You need to process t test cases.
The first line contains an integer t (1 β€ t β€ 3025) β the number of test cases. Then descriptions of t test cases follow.
The first line of each test case contains four integers a, b, c, d (0 β€ a,\,b,\,c,\,d β€ 9; d β€ a; b β€ c).
One can show that for any test case satisfying the constraints above, there is at least one olympiad scenario possible.
Output
For each test case print a single integer β the smallest possible cutoff score in some olympiad scenario satisfying the given information.
Example
Input
2
1 2 2 1
4 8 9 2
Output
3
12
Note
For the first test case, consider the following olympiad scenario: there are 101 participants in the elimination stage, each having 1 point for the first contest and 2 points for the second contest. Hence the total score of the participant on the 100-th place is 3.
For the second test case, consider the following olympiad scenario:
* there are 50 participants with points 5 and 9 for the first and second contest respectively;
* 50 participants with points 4 and 8 for the first and second contest respectively;
* and 50 participants with points 2 and 9 for the first and second contest respectively.
Hence the total point score of the participant on the 100-th place is 12.
Submitted Solution:
```
for i in range(int(input())):
l=list(map(int,input().split()))
print(l[0]+l[1])
```
No
| 94,599 |
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